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One dimensional motion with constant acceleration

Summary: The motion on earth is often modified with constant acceleration due to the combination of gravity and friction forces.

Free falling bodies under gravity represents typical case of motion in one dimension with constant acceleration. A body projected vertically upwards is also a case of constant acceleration in one dimension, but with the difference that body undergoes reversal of direction as well after reaching the maximum height. Yet another set of examples of constant accelerations may include object sliding on an incline plane, motion of an aboject impeded by rough surfaces and many other motions under the influence of gravitational and frictional forces.

The defining differential equations of velocity and acceleration involve only one position variable (say x). In the case of motion under constant acceleration, the differential equation defining acceleration must evaluate to a constant value.

v = đ x đ t i v = đ x đ t i

and

a = đ 2 x đ t 2 i = k i a = đ 2 x đ t 2 i = k i

where k is a positive or negative constant.

The corresponding scalar form of the defining equations of velocity and acceleration for one dimensional motion with constant acceleration are :

v = đ x đ t v = đ x đ t

and

a = đ 2 x đ t 2 = k a = đ 2 x đ t 2 = k

Example 1: Constant acceleration

Problem : The position “x” in meter of a particle moving in one dimension is described by the equation :

t = x + 1 t = x + 1

where “t” is in second.

Find the time when velocity is zero.
Does the velocity changes its direction?
Locate position of the particle in the successive seconds for first 3 seconds.
Find the displacement of the particle in first three seconds.
Find the distance of the particle in first three seconds.
Find the displacement of the particle when the velocity becomes zero.
Determine, whether the particle is under constant or variable force.

Solution : Velocity is equal to the first differential of the position with respect to time, while acceleration is equal to the second differential of the position with respect to time. The given equation, however, expresses time, t, in terms of position, x. Hence, we need to obtain expression of position as a function in time.

t = x + 1 ⇒ x = t - 1 t = x + 1 ⇒ x = t - 1

Squaring both sides, we have :

⇒ x = t 2 - 2 t + 1 ⇒ x = t 2 - 2 t + 1

This is the desired expression to work upon. Now, taking first differential w.r.t time, we have :

v = đ x đ t = đ đ t ( t 2 - 2 t + 1 ) = 2 t - 2 v = đ x đ t = đ đ t ( t 2 - 2 t + 1 ) = 2 t - 2

1: When v = 0, we have v = 2t – 2 = 0

⇒ t = 1 s ⇒ t = 1 s

2. Velocity is expressed in terms of time as :

v = 2 t - 2 v = 2 t - 2

It is clear from the expression that velocity is negative for t < 1 second, while positive for t > 1. As such velocity changes its direction during motion.

3: Positions of the particle at successive seconds for first three seconds are :

t = 0 ; x = t 2 - 2 t + 1 = 0 - 0 + 1 = 1 m t = 1 ; x = t 2 - 2 t + 1 = 1 - 2 + 1 = 0 m t = 2 ; x = t 2 - 2 t + 1 = 4 - 4 + 1 = 1 m t = 3 ; x = t 2 - 2 t + 1 = 9 - 6 + 1 = 4 m t = 0 ; x = t 2 - 2 t + 1 = 0 - 0 + 1 = 1 m t = 1 ; x = t 2 - 2 t + 1 = 1 - 2 + 1 = 0 m t = 2 ; x = t 2 - 2 t + 1 = 4 - 4 + 1 = 1 m t = 3 ; x = t 2 - 2 t + 1 = 9 - 6 + 1 = 4 m

Figure 1
Graphical representation of position

4: Positions of the particle at t = 0 and t = 3 s are 1 m and 4 m from the origin.

Hence, displacement in first three seconds is 4 – 1 = 3 m

5: The particle moves from the start position, x = 1 m, in the negative direction for 1 second. At t = 1, the particle comes to rest. For the time interval from 1 to 3 seconds, the particle moves in the positive direction.

Distance in the interval t = 0 to 1 s is :

s 1 = 1 - 0 = 1 m s 2 = 4 - 0 = 4 m s 1 = 1 - 0 = 1 m s 2 = 4 - 0 = 4 m

Total distance is 1 + 4 = 5 m.

6: Velocity is zero, when t = 1 s. In this period, displacement is 1 m.

7: In order to determine the nature of force on the particle, we first determine the acceleration as :

a = đ v đ t = đ đ t ( 2 t - 2 ) = 2 m / s 2 a = đ v đ t = đ đ t ( 2 t - 2 ) = 2 m / s 2

Acceleration of the motion is constant, independent of time. Hence, force on the particle is also constant during the motion.

Equation of motion for one dimensional motion with constant acceleration

The equation of motions in one dimension for constant acceleration is obtained from the equations of motion established for the general case i.e. for the three dimensional motion. In one dimension, the equation of motion is simplified (r is replaced by x or y or z with corresponding unit vector). The three basic equations of motions are (say in x - direction) :

1: v = u + a t 2: v avg = ( u + v ) 2 3: Δ x i = u t + 1 2 a t 2 1: v = u + a t 2: v avg = ( u + v ) 2 3: Δ x i = u t + 1 2 a t 2

Significantly, we can treat vector equivalently as scalars with appropriate sign. Following is the construct used for this purpose :

Sign convention

Assign an axis along the motion. Treat direction of axis as positive.
Assign the origin with the start of motion or start of observation. It is, however, a matter of convenience and is not a requirement of the construct.
Use all quantities describing motion in the direction of axis as positive.
Use all quantities describing motion in the opposite direction of axis as negative.

Once, we follow the rules as above, we can treat equations of motion as scalar equations as :

1: v = u + a t 2: v avg = ( u + v ) 2 3: Δ x = u t + 1 2 a t 2 1: v = u + a t 2: v avg = ( u + v ) 2 3: Δ x = u t + 1 2 a t 2

Example 2: Constant acceleration

Problem : A car moving with constant acceleration covers two successive kilometers in 20 s and 30 s respectively. Find the acceleration of the car.

Solution : Let "u" and "a" be the initial velocity and acceleration of the car. Applying third equation of motion for first kilometer, we have :

1000 = u x 20 + 1 2 a 20 2 = 20 u + 200 a ⇒ 100 = 2 u + 20 a 1000 = u x 20 + 1 2 a 20 2 = 20 u + 200 a ⇒ 100 = 2 u + 20 a

At the end of second kilometer, total displacement is 2 kilometer (=2000 m) and total time is 20 + 30 = 50 s. Again applying third equation of motion,

2000 = u x 50 + 1 2 a 50 2 = 50 u + 2500 a ⇒ 200 = 5 u + 125 a 2000 = u x 50 + 1 2 a 50 2 = 50 u + 2500 a ⇒ 200 = 5 u + 125 a

Solving two equations,

a = - 2.86 m / s 2 a = - 2.86 m / s 2

Note that acceleration is negative to the positive direction (direction of velocity) and as such it is termed “deceleration”. This interpretation is valid as we observe that the car covers second kilometer in longer time that for the first kilometer, which means that the car has slowed down.

It is important to emphasize here that mere negative value of acceleration does not mean it to be deceleration. The deciding criterion for deceleration is that acceleration should be opposite to the direction of velocity.

Motion under gravity

We have observed that when a feather and an iron ball are released from a height, they reach earth surface with different velocity and at different times. These objects are under the action of different forces like gravity, friction, wind and buoyancy force. In case forces other than gravity are absent like in vacuum, the bodies are only acted by the gravitational pull towards earth. In such situation, acceleration due to gravity, denoted by g, is the only acceleration.

The acceleration due to gravity near the earth surface is nearly constant and equal to 9.8 m / s 2 m / s 2 . Value of ‘g’ is taken as 10 m / s 2 m / s 2 as an approximation to facilitate ease of calculation.

When only acceleration due to gravity is considered, neglecting other forces, each of the bodies (feather and iron ball) starting from rest is accelerated at the same rate. Velocity of each bodies increases by 9.8 m/s at the end of every second. As such, the feather and the iron ball reach the surface at the same time and at the same velocity.

Additional equations of motion

A close scrutiny of three equations of motion derived so far reveals that they relate specific quantities, which define the motion. There is possibility that we may encounter problems where inputs are not provided in the manner required by equation of motion.

For example, a problem involvoing calaculation of displacement may identify initial velocity, final velocity and acceleration as input. Now, the equation of motion for displacement is expressed in terms of initial velocity, time and acceleration. Evidently, there is a mis-match between what is given and what is required. We can, no doubt, find out time from the set of given inputs, using equation for velocity and then we can solve equation for the displacement. But what if we develop a relation-ship which relates the quantities as given in the input set! This would certainly help.

From first equation :

v = u + a t ⇒ v - u = a t ⇒ t = ( v - u ) a = a t v = u + a t ⇒ v - u = a t ⇒ t = ( v - u ) a = a t

From second equation :

( u + v ) 2 = ( x 2 - x 1 ) t ⇒ ( u + v ) = 2 ( x 2 - x 1 ) t ( u + v ) 2 = ( x 2 - x 1 ) t ⇒ ( u + v ) = 2 ( x 2 - x 1 ) t

Eliminating ‘t’,

⇒ ( u + v ) = 2 a ( x 2 - x 1 ) ( v - u ) ⇒ v 2 - u 2 = 2 a ( x 2 - x 1 ) ⇒ ( u + v ) = 2 a ( x 2 - x 1 ) ( v - u ) ⇒ v 2 - u 2 = 2 a ( x 2 - x 1 )

⇒ v 2 - u 2 = 2 a ( x 2 - x 1 )

(1)

This equation relates initial velocity, final velocity, acceleration and displacement.

Also, we observe that equation for displacement calculates displacement when initial velocity, acceleration and time are given. If final velocity - instead of initial velocity - is given, then displacement can be obtained with slight modification.

Δ x = x 2 - x 1 = u t = 1 2 a t 2 Δ x = x 2 - x 1 = u t = 1 2 a t 2

Using v = u + at,

⇒ Δ x = x 2 - x 1 = ( v - a t ) t + 1 2 a t 2 ⇒ Δ x = x 2 - x 1 = ( v - a t ) t + 1 2 a t 2

⇒ Δ x = x 2 - x 1 = v t - 1 2 a t 2

(2)

Example 3: Constant acceleration in one dimensional motion

Problem : A train traveling on a straight track moves with a speed of 20 m/s. Brake is applied uniformly such that its speed is reduced to 10 m/s, while covering a distance of 200 m. With the same rate of deceleration, how far will the train go before coming to rest.

Solution : To know the distance before train stops, we need to know the deceleration. We can find out deceleration from the first set of data. Here, u = 20 m/s ; v = 10 m/s ; x = 200 m ; a = ? An inspection of above data reveals that none of the first three equations of motion fits the requirement in hand, while the additional form of equation v 2 - u 2 = 2 a x v 2 - u 2 = 2 a x serves the purpose :

a = v 2 - u 2 2 x = 10 2 - 20 2 2 x 200 = - 3 4 m / s 2 a = v 2 - u 2 2 x = 10 2 - 20 2 2 x 200 = - 3 4 m / s 2

For the train to stops, we have

u = 20 m / s ; v = 0 m / s ; a = - 3 4 m / s 2 ; x = ? u = 20 m / s ; v = 0 m / s ; a = - 3 4 m / s 2 ; x = ?

and,

x = v 2 - u 2 2 a = 0 2 - 20 2 2 x ( - 3 4 ) = 266.67 m x = v 2 - u 2 2 a = 0 2 - 20 2 2 x ( - 3 4 ) = 266.67 m

Displacement in a particular second

The displacement in a second is obtained by subtracting two displacements in successive seconds. We calculate displacements in n th n th second and (n-1) th (n-1) th seconds. The difference of two displacements is the displacement in the n th n th second. Now, the displacements at the end of n and (n-1) seconds as measured from origin are given by :

x n = u n + 1 2 a n 2 x n - 1 = u ( n - 1 ) + 1 2 a ( n - 1 ) 2 x n = u n + 1 2 a n 2 x n - 1 = u ( n - 1 ) + 1 2 a ( n - 1 ) 2

The displacement in the n th n th second, therefore, is :

s n = x 2 - x 1 = u n + 1 2 a n 2 - u n + u - 1 2 a n 2 - 1 2 a + a n ⇒ s n = x 2 - x 1 = u - 1 2 a + a n s n = x 2 - x 1 = u n + 1 2 a n 2 - u n + u - 1 2 a n 2 - 1 2 a + a n ⇒ s n = x 2 - x 1 = u - 1 2 a + a n

⇒ s n = x 2 - x 1 = u + a 2 ( 2 n - 1 )

(3)

Example 4

Problem : The equation of motion for displacement in n th n th second is given by :

x n = u + a 2 ( 2 n - 1 ) x n = u + a 2 ( 2 n - 1 )

This equation is dimensionally incompatible, yet correct. Explain.

Solution : Since “n” is a number, the dimension of the terms of the equation is indicated as :

[ x n ] = [ Displacment ] = [ L ] [ u ] = [ velocity ] = [ L T - 1 ] [ a 2 ( 2 n - 1 ) ] = [ acceleration ] = [ L T - 2 ] [ x n ] = [ Displacment ] = [ L ] [ u ] = [ velocity ] = [ L T - 1 ] [ a 2 ( 2 n - 1 ) ] = [ acceleration ] = [ L T - 2 ]

Clearly, dimensions of the terms are not same and hence equation is apparently incompatible in terms of dimensions.

In order to resolve this apparent incompatibility, we need to show that each term of right hand side of the equation has dimension that of displacement i.e. length. Now, we know that the relation is derived for a displacement for time equal to “1” second. As multiplication of “1” or " 1 2 1 2 " with any term is not visible, the apparent discrepancy has appeared in otherwise correct equation. We can, therefore, re-write the equation with explicit time as :

x n = u x 1 + a 2 ( 2 n - 1 ) x 1 2 ⇒ x n = u x t + a 2 ( 2 n - 1 ) x t 2 x n = u x 1 + a 2 ( 2 n - 1 ) x 1 2 ⇒ x n = u x t + a 2 ( 2 n - 1 ) x t 2

Now, the dimensions of the first and second terms on the right side are :

[ u x t ] = [ L T - 1 T ] = [ L ] [ a 2 ( 2 n - 1 ) x t 2 ] = [ L T - 2 T 2 ] = [ L ] [ u x t ] = [ L T - 1 T ] = [ L ] [ a 2 ( 2 n - 1 ) x t 2 ] = [ L T - 2 T 2 ] = [ L ]

Thus, we see that the equation is implicitly correct in terms of dimensions.

Exercise 1

If a particle, moving in straight line, covers distances “x”, “y” and “z” in p th p th , q th q th and r th r th seconds respectively, then prove that :

( p - q ) z + ( r - p ) y + ( q - r ) x = 0 ( p - q ) z + ( r - p ) y + ( q - r ) x = 0

Solution 1

This is a motion with constant acceleration in one dimension. Let the particle moves with a constant acceleration, “a”. Now, the distances in particular seconds as given in the question are :

x = u + a 2 ( 2 p - 1 ) y = u + a 2 ( 2 q - 1 ) z = u + a 2 ( 2 r - 1 ) x = u + a 2 ( 2 p - 1 ) y = u + a 2 ( 2 q - 1 ) z = u + a 2 ( 2 r - 1 )

Subtracting, second from first equation,

x - y = a 2 ( 2 p - 1 - 2 q + 1 ) = a ( p - q ) ⇒ ( p - q ) = ( x - y ) a ⇒ ( p - q ) z = ( x - y ) z a x - y = a 2 ( 2 p - 1 - 2 q + 1 ) = a ( p - q ) ⇒ ( p - q ) = ( x - y ) a ⇒ ( p - q ) z = ( x - y ) z a

Similarly,

⇒ ( r - p ) y = ( z - x ) y a ⇒ ( q - r ) x = ( y - z ) x a ⇒ ( r - p ) y = ( z - x ) y a ⇒ ( q - r ) x = ( y - z ) x a

Adding three equations,

( p - q ) z + ( r - p ) y + ( q - r ) x = 1 a ( x y - y z + y z - x y + x y - x z ) ⇒ ( p - q ) z + ( r - p ) y + ( q - r ) x = 0 ( p - q ) z + ( r - p ) y + ( q - r ) x = 1 a ( x y - y z + y z - x y + x y - x z ) ⇒ ( p - q ) z + ( r - p ) y + ( q - r ) x = 0

Average acceleration

Average acceleration is ratio of change in velocity and time. This is an useful concept where acceleration is not constant throughout the motion. There may be motion, which has constant but different values of acceleration in different segments of motion. Our job is to find an equivalent constant acceleration, which may be used to determine attributes for the whole of motion. Clearly, the single value equivalent acceleration should be such that it yields same value of displacement and velocity for the entire motion. This is the underlying principle for determining equivalent or average acceleration for the motion.

Example 5

Problem : A particle starting with velocity “u” covers two equal distances in a straight line with accelerations a 1 a 1 and a 2 a 2 . What is the equivalent acceleration for the complete motion?

Solution : The equivalent acceleration for the complete motion should yield same value for the attributes of the motion at the end of the journey. For example, the final velocity at the end of the journey with the equivalent acceleration should be same as the one calculated with individual accelerations.

Here initial velocity is "u". The velocity at the end of half of the distance (say “x”) is :

v 1 2 = u 2 + 2 a 1 x v 1 2 = u 2 + 2 a 1 x

Clearly, v 1 v 1 is the initial velocity for the second leg of motion,

v 2 2 = v 1 2 + 2 a 2 x v 2 2 = v 1 2 + 2 a 2 x

Adding above two equations, we have :

v 2 2 = u 2 + 2 ( a 1 + a 2 ) x v 2 2 = u 2 + 2 ( a 1 + a 2 ) x

This is the square of velocity at the end of journey. Now, let “a” be the equivalent acceleration, then applying equation of motion for the whole distance (2x),

v 2 2 = u 2 + 2 a ( 2 x ) v 2 2 = u 2 + 2 a ( 2 x )

Comparing equations,

a = a 1 + a 2 2 a = a 1 + a 2 2

Interpretation of equations of motion

One dimensional motion felicitates simplified paradigm for interpreting equations of motion. Description of motion in one dimension involves mostly the issue of “magnitude” and only one aspect of direction. The only possible issue of direction here is that the body undergoing motion in one dimension may reverse its direction during the course of motion. This means that the body may either keep moving in the direction of initial velocity or may start moving in the opposite direction of the initial velocity at certain point of time during the motion. This depends on the relative direction of initial velocity and acceleration. Thus, there are two paradigms :

Constant force is applied in the direction of initial velocity.
Constant force is applied in the opposite direction of initial velocity.

Irrespective of the above possibilities, one fundamental attribute of motion in one dimension is that all parameters defining motion i.e initial velocity, final velocity and acceleration act along a straight line.

Constant acceleration (force) is applied in the direction of velocity

The magnitude of velocity increases by the magnitude of acceleration at the end of every second (unit time interval). In this case, final velocity at any time instant is greater than velocity at an earlier instant. The motion is not only in one dimension i.e. linear , but also unidirectional. Take the example of a ball released (initial velocity is zero) at a certain height ‘h’ from the surface. The velocity of the ball increases by the magnitude of ‘g’ at the end of every second. If the body has traveled for 3 seconds, then the velocity after 3 seconds is 3g (v= 0 + 3 x g = 3g m/s).

Figure 2: Attributes of motion as the ball falls under gravity
Attributes of motion

In this case, all parameters defining motion i.e initial velocity, final velocity and acceleration not only act along a straight line, but also in the same direction. As a consequence, displacement is always increasing during the motion like distance. This fact results in one of the interesting aspect of the motion that magnitude of displacement is equal to distance. For this reason, average speed is also equal to the magnitude of average velocity.

s = | x | s = | x |

and

Δ s Δ t = | Δ x Δ t | Δ s Δ t = | Δ x Δ t |

Constant acceleration (force) is applied in the opposite direction of velocity

The magnitude of velocity decreases by the magnitude of acceleration at the end of every second (unit time interval). In this case, final velocity at any time instant is either less than velocity at an earlier instant or has reversed its direction. The motion is in one dimension i.e. linear, but may be unidirectional or bidirectional. Take the example of a ball thrown (initial velocity is ,say, 30 m/s) vertically from the surface. The velocity of the ball decreases by the magnitude of ‘g’ at the end of every second. If the body has traveled for 3 seconds, then the velocity after 3 seconds is 30 - 3g = 0 (assume g = 10 m / s 2 m / s 2 ).

Figure 3: Attributes of motion as the ball moves up against gravity
Attributes of motion

During upward motion, velocity and acceleration due to gravity are in opposite direction. As a result, velocity decreases till it achieves the terminal velocity of zero at the end of 3rd second. Note that displacement during the motion is increasing till the ball reaches the maximum height.

At the maximum height, the velocity of the ball is zero and is under the action of force due to gravity as always during the motion. As such, the ball begins moving in downward direction with the acceleration due to gravity. The directions of velocity and acceleration, in this part of motion, are same. Note that displacement with respect to point of projection is decreasing.

In the overall analysis of motion when initial velocity is against acceleration, parameters defining motion i.e initial velocity, final velocity and acceleration act along a straight line, but in different directions. As a consequence, displacement may either be increasing or decreasing during the motion. This means that magnitude of displacement may not be equal to distance. For example, consider the motion of ball from the point of projection, A, to maximum height, B, to point, C, at the end of 4 seconds. The displacement is 40 m, while distance is 45 + 5 = 50 m as shown in the figure below.

Figure 4
Attributes of motion

For this reason, average speed is not always equal to the magnitude of average velocity.

s ≠ | x | s ≠ | x |

and

Δ s Δ t ≠ | Δ x Δ t | Δ s Δ t ≠ | Δ x Δ t |

Exercises

Exercise 2

Two cyclists start off a race with initial velocities 2 m/s and 4 m/s respectively. Their linear accelerations are 2 and 1 m / s 2 m / s 2 respectively. If they reach the finish line simultaneously, then what is the length of the track?

Solution 2

This is a case of one dimensional motion with constant acceleration. Since both cyclists cross the finish line simultaneously, they cover same displacement in equal times. Hence,

x 1 = x 2 x 1 = x 2

u 1 t + 1 2 a 1 t 2 = u 2 t + 1 2 a 2 t 2 u 1 t + 1 2 a 1 t 2 = u 2 t + 1 2 a 2 t 2

Putting values as given in the question, we have :

2 t + 1 2 x 2 x t 2 = 4 t + 1 2 x 1 x t 2 ⇒ t 2 - 4 t = 0 ⇒ t = 0 s , 4 s 2 t + 1 2 x 2 x t 2 = 4 t + 1 2 x 1 x t 2 ⇒ t 2 - 4 t = 0 ⇒ t = 0 s , 4 s

Neglecting zero value,

⇒ t = 4 s ⇒ t = 4 s

The linear distance covered by the cyclist is obtained by evaluating the equation of displacement of any of the cyclists as :

x = 2 t + 1 2 x 2 x t 2 = 2 x 4 + 1 2 x 2 x 4 2 = 24 t x = 2 t + 1 2 x 2 x t 2 = 2 x 4 + 1 2 x 2 x 4 2 = 24 t

Exercise 3

Two cars are flagged off from the starting point. They move with accelerations a 1 a 1 and a 2 a 2 respectively. The car “A” takes time “t” less than car “B” to reach the end point and passes the end point with a difference of speed, “v”, with respect to car “B”. Find the ratio v/t.

Solution 3

Both cars start from rest. They move with different accelerations and hence take different times to reach equal distance, say t 1 t 1 and t 2 t 2 for cars A and B respectively. Their final speeds at the end points are also different, say v 1 v 1 and v 2 v 2 for cars A and B respectively. According to the question, the difference of time is “t”, whereas difference of speeds is “v”.

As car A is faster, it takes lesser time. Here, t 2 > t 1 t 2 > t 1 . The difference of time, “t”, is :

t = t 2 - t 1 t = t 2 - t 1

From equation of motion,

x = 1 2 a 1 t 1 2 ⇒ t 1 = √ ( 2 x a 1 ) x = 1 2 a 1 t 1 2 ⇒ t 1 = √ ( 2 x a 1 )

Similarly,

⇒ t 2 = √ ( 2 x a 2 ) ⇒ t 2 = √ ( 2 x a 2 )

Hence,

⇒ t = t 2 - t 1 = √ ( 2 x a 2 ) - √ ( 2 x a 1 ) ⇒ t = t 2 - t 1 = √ ( 2 x a 2 ) - √ ( 2 x a 1 )

Car A is faster. Hence, v 1 > v 2 v 1 > v 2 . The difference of time, “v”, is :

⇒ v = v 1 - v 2 ⇒ v = v 1 - v 2