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)

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

not moving. pos accel,