Question

(4. A bus is moving at 25 m/s when the driver steps on the brakes and brings the bus to a stop in
3.0 s.
a. What is the average acceleration of the bus while braking?
b. If the bus took twice as long to stop, how would the acceleration compare with what you
found in part a?​

Answer 1

Answer:

(a)  the average acceleration of the bus while braking is 8.333 m/s

(b) if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.  [¹/₂ (8.333 m/s) = 4.16 s]

Explanation:

Given;

initial velocity of the bus, v = 25 m/s

time of the motion, t = 3 s

(a)  the average acceleration of the bus while braking

a = dv/dt

where;

a is the bus acceleration

dv is change in velocity

dt is change in time

a = 25 / 3

a = 8.333 m/s

(b) If the bus took twice as long to stop, the duration = 2 x 3s

a = 25 / (2 x 3s)

a = ¹/₂ x (25 / 3)

a = ¹/₂ (8.333 m/s) = 4.16 s

Thus, if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.

Answer 2

(a) The average acceleration of the bus while braking is 8.333 m/s.

(b) In the case when the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.  [¹/₂ of 8.333 m/s) = 4.16 s]

Calculation of acceleration:

initial velocity of the bus, v = 25 m/s

time of the motion, t = 3 s

(a)  the average acceleration of the bus should be

a = dv/dt

Here,

a is the bus acceleration

DV is a change in velocity

dt is changed in time

So,

a = 25 / 3

a = 8.333 m/s

(b) If the bus took twice as long to stop, the duration = 2 x 3s

So,

a = 25 / (2 x 3s)

a = ¹/₂ x (25 / 3)

a = ¹/₂ (8.333 m/s)

= 4.16 s

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