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.
Answer:
(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:
<em>v = u + (a)(t)
</em>
where <em>v </em>is final velocity, <em>u </em>is initial velocity, <em>a </em>is acceleration and <em>t </em>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
<em>2(a)(s) = v² – u²</em>
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)
