A car sits in 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 19 m/s when it reaches the end of the 110-m-long ramp. The traffic on the freeway is moving at a constant speed of 19 m/s. What distance does the traffic travel while the car is moving the length of the ramp

the distance travelled by the traffic, while the car is moving the length of the ramp is 220 m

Explanation:

Given the data in the question;

first we determine acceleration  using the kinematic equation below;

v² - u² = 2as

a = v² - u² / 2s

our initial velocity is zero, v is 19 m/s and distance s is 110 m

so we substitute

a = (19² - 0²) / 2×110

a = 361 / 220

a = 1.6409 m/s²

Next, the time t taken by the car to travel along the length of the ramp will be;

t = v - u / a

we substitute

t = (19 - 0) / 1.6409

t = 11.579 sec

so the distance travelled by a body moving with constant speed u is given by following expression:

s = ut + at²

so we substitute 19 m/s for u, 11.579 sec for t and 0 m/s² for a

s = (19 m/s × 11.579 s) + × 0 × (11.579)²

s = (19 m/s × 11.579 s) + 0

s = 220 m

Therefore, the distance travelled by the traffic, while the car is moving the length of the ramp is 220 m

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