A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver 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 20 m/s (45 mi/h) when it reaches the end of the 120-m-long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of 20 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

a) As we have as givens the final speed (and the initial also, which is just 0), and the distance travelled, as the acceleration is constant, we can use the following kinematic equation in order to find the acceleration:

[tex]vf^{2} -vo^{2} = 2*a*x[/tex]

Replacing vf = 20 m/s, and x= 120m, and solving for a, we get:

[tex]a=\frac{vf^{2}}{2*x} = \frac{(20m/s)^{2} }{2*120m} = 1.7 m/s2[/tex]

b) We can find the time just applying the definition of average acceleration and rearranging terms, as follows:

vf = v₀ + a*t = a*t

Replacing by vf= 20 m/s and a = 1.7 m/s², we can solve for t, as follows:

[tex]t =\frac{vf}{a} = \frac{20m/s}{1.7m/s2} = 12 s[/tex]

c) As the traffic is moving at constant speed of 20 m/s, we can find the distance travelled just applying the definition of average velocity and rearranging terms, as follows:

Δx = vₓ*t = 20 m/s*12 s = 240 m