A minivan is tested for acceleration and braking. In the street-start acceleration test, the elapsed time is 8.6 s for a velocity increase from 10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m during braking to a stop from 100 km/h. Assume constant values of acceleration and deceleration.

The question is incomplete. Here is the complete question:

A minivan is tested for acceleration and braking. In the street-start acceleration test, the elapsed time is 8.6 s for a velocity increase from 10 km/h to 100 km/h. In the braking test, the distance traveled is 44 m during braking to a stop from 100 km/h. Assume constant values of acceleration and deceleration. Determine  

(a) the acceleration during the street-start test,  

(b) the deceleration during the braking test.

(a)37500 km/h²

(b)113636.36 km/h²

Explanation:

part (a)

Because it is given that we can assume constant acceleration therefore we can use the following equation of motion:

v = u + (a)(t)

where v is final velocity, u is initial velocity, a is acceleration and t is time change

Given in the question:

v = 100km/h

u = 10 km/h

t = 8.6 sec (changing to hours)

t = 0.0024 hours (round off to 4 decimal places)

100 = 10 + (a x 0.0024)

Rearranging the equation to find value of a

a = (100 – 10) / 0.0024

a = 37500 km/h² (Answer)

part (b)

Now we can use the following equation to find deceleration

2(a)(s) = v² – u²

Where a is acceleration, s is distance travelled, v is final velocity and u is initial velocity

Given in the question

s = 44 m

changing to km

s = 0.044 km

v = 0 km/h (because it stops)

u = 100 km/h

2(a)(0.044) = (0)² – (100)²

0.088(a) = 0 - 10000  

a = - 10000/0.088

a = - 113636.36 km/h2  

The negative sign in the answer shows that it is deceleration

Therefore deceleration = 113636.36 km/h² (Answer)

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