A car 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