Problem : At a certain instant, the components of velocity and acceleration are given as :
v
x
=
4
m
/
s
;
v
y
=
3
m
/
s
;
a
x
=
2
m
/
s
2
;
a
y
=
1
m
/
s
2
.
v
x
=
4
m
/
s
;
v
y
=
3
m
/
s
;
a
x
=
2
m
/
s
2
;
a
y
=
1
m
/
s
2
.
What is the rate of change of speed?
Solution : Here, phrasing of question is important. We are required to find the rate of change of speed – not the rate of change of velocity or magnitude of rate of change of velocity. Let us have a look at the subtle differences in the meaning here :
1: The rate of change of velocity
The rate of change of velocity is equal to acceleration. For the given two dimensional motion,
a
=
đ
v
đ
t
=
đ
đ
t
(
a
x
i
+
a
y
j
)
a
=
đ
v
đ
t
=
2
i
+
1
j
a
=
đ
v
đ
t
=
đ
đ
t
(
a
x
i
+
a
y
j
)
a
=
đ
v
đ
t
=
2
i
+
1
j
2: The magnitude of rate of change of velocity
The magnitude of rate of change of velocity is equal to magnitude of acceleration. For the given two dimensional motion,

a

=

đ
v
đ
t

=
√
(
2
2
+
1
2
)
=
√
5
m
/
s
2

a

=

đ
v
đ
t

=
√
(
2
2
+
1
2
)
=
√
5
m
/
s
2
3: The rate of change of speed
The rate of change of speed (dv/dt) is not equal to the magnitude of acceleration, which is equal to the absolute value of the rate of change of velocity. It is so because speed is devoid of direction, whereas acceleration consists of both magnitude and direction.
Let “v” be the instantaneous speed, which is given in terms of its component as :
v
2
=
v
x
2
+
v
y
2
v
2
=
v
x
2
+
v
y
2
We need to find rate of change of speed i.e dv/dt, using the values given in the question. Therefore, we need to differentiate speed with respect to time,
⇒
2
v
đ
v
đ
t
=
2
v
x
(
đ
v
x
đ
t
)
+
2
v
y
(
đ
v
y
đ
t
)
⇒
2
v
đ
v
đ
t
=
2
v
x
(
đ
v
x
đ
t
)
+
2
v
y
(
đ
v
y
đ
t
)
If we ponder a bit, then we would realize that when we deal with component speed or magnitude of component velocity then we are essentially dealing with unidirectional motion. No change in direction is possible as components are aligned to a fixed axis. As such, equating rate of change in speed with the magnitude of acceleration in component direction is valid. Now, proceeding ahead,
⇒
đ
v
đ
t
=
v
x
a
x
+
v
y
a
y
v
⇒
đ
v
đ
t
=
v
x
a
x
+
v
y
a
y
v
⇒
đ
v
đ
t
=
v
x
a
x
+
v
y
a
y
√
(
v
x
2
+
v
y
2
)
⇒
đ
v
đ
t
=
v
x
a
x
+
v
y
a
y
√
(
v
x
2
+
v
y
2
)
Putting values, we have :
⇒
đ
v
đ
t
=
4
x
2
+
3
x
1
√
(
4
2
+
3
2
)
⇒
đ
v
đ
t
=
11
5
=
2.2
m
/
s
⇒
đ
v
đ
t
=
4
x
2
+
3
x
1
√
(
4
2
+
3
2
)
⇒
đ
v
đ
t
=
11
5
=
2.2
m
/
s
Note: This is an important question as it brings out differences in interpretation of familiar terms. In order to emphasize the difference, we summarize the discussion as hereunder :
i: In general (i.e two or three dimensions),
⇒
đ
v
đ
t
≠

đ
v
đ
t

⇒
đ
v
đ
t
≠

a

⇒
đ
v
đ
t
≠
a
⇒
đ
v
đ
t
≠

đ
v
đ
t

⇒
đ
v
đ
t
≠

a

⇒
đ
v
đ
t
≠
a
ii: In the case of one dimensional motion, the inequality as above disappears.
⇒
đ
v
đ
t
=
a
⇒
đ
v
đ
t
=
a