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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

Sagot :

nuhulawal20

Answer:

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 + [tex]\frac{1}{2}[/tex] 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) + [tex]\frac{1}{2}[/tex] × 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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