Mr. Sharma brakes his car with constant acceleration from a velocity of 25 m/s to 15 m/s over a
distance of 200 m.
(a) How much time elapses during this interval?
(b) What is the acceleration?
If he has to continue braking with the same constant acceleration, how much longer
would it take for him to stop and how much additional distance would he cover?
Answers
Answer:
We select positive direction for our coordinate system to be the direction of the velocity and choose
the origin so that xi
= 0 when the braking begins. Then the initial velocity is ux
= +25 m/s at t = 0, and
the final velocity and position are vx
= +15 m/s and x = 200 m at time t.
Since the acceleration is constant, the average velocity in the interval can be found from the average
of the initial and final velocities.
∴ vav, x
= 2
1
(ux
+ vx
) = 2
1
(15 + 25) = 20 m/s.
The average velocity can also be expressed as vav, x
=
t
x
∆
∆
. With ∆x = 200 m and ∆t = t − 0,
we can solve for t:
t = vav,x
∆x
= 20
200
= 10 s.
(b) We can now find the acceleration using vx
= ux
+ ax
t
ax
= t
vx − ux
= 10
15 − 25
= − 1 m/s2
.
The acceleration is negative, which means that the positive velocity is becoming smaller as brakes
are applied (as expected).
(c) Now with known acceleration, we can find the total time for the car to go from velocity ux
= 25 m/s to
vx
= 0. Solving for t, we find
t =
x
x x
a
v − u
= 1
0 25
−
−
= 25 s.
The total distance covered is
x = xi
+ ux
t + 2
1
ax
t
2
= 0 + (25)(25) + 2
1
(−1)(25)2
= 625 − 312.5 = 312.5 m.
Additional distance covered = 312.5 – 200 = 112.5 m
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