Acar sits in an entrance ramp to a freeway, waiting for a break in the traffic. the driver sees a small gap between a van and an 18-wheel truck and accelerates with constant acceleration along the ramp and onto the freeway. the car starts from rest, moves in a straight line, and has a speed of 16.0 m/s when it reaches the end of the ramp, which has length 125 m . what is the acceleration of the car?
how much time does it take the car to travel the length of the ramp?

a=1.024m/s

t=15.62s

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t                         (1)

{Vf^{2}-Vo^2}/{2.a} =X      (2)

X=Xo+ VoT+0.5at^{2}      (3)

X=(Vf+Vo)T/2                   (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 4 above equations and use algebra to solve

for this problem

Vf=16m/s

Vo=0m/s, the cart starts from the rest

X=125m

we can use the ecuation number tow to calculate the acceleration

{Vf^{2}-Vo^2}/{2.a} =X

{Vf^{2}-Vo^2}/{2.x} =a

{16^{2}-0^2}/{2(125)} =a

a=1.024m/s

to calculate the time we can use the ecuation number 1

Vf=Vo+a.t    

t=(Vf-Vo)/a

t=(16-0)/1.024

t=15.62s

i would say