A person driving her car at 45 km/h approaches an intersection justas the traffi

Question

A person driving her car at 45 km/h approaches an intersection justas the traffic light turns yellow. She knows that the yellow lightlasts only 2.0s before turning red, and she is 28 m away from thenear side of the intersection. Should she try to stop or should shespeed up the cross the intersection before the light turns red? Theintersection is 15 m wide. Her car's maximum deceleration is -5.8m/s^2 whereas it can accelerate from 45 km/h to 65km/h in 6/0s.Ignore the lengh of her car and her reaction time.

Explanation / Answer

Part 1: Braking in time If this person breaks, then by KE = braking PE, we have    1/2 m v^2 = m *a * s, where m is the mass of thevehicle and passenger, v is its velocity and s is the stoppingdistance. So, v^2 = 2 * a *s or s = v^2/(2 *a) = (45 * 1000/ 3600m/s)^2 /(2 * 5.8 m/s^2) = 13.47 m << 28 m So, stopping would work. Part 2: Accelerating Distance to cover is : 28m + 15 m = 43 m Acceleration is (65-45)/6 = 20/6 = 10.3 = 3.33 m/s^2 Calculate distance for speed + acceleration in 2 secondsin 45*1000/3600 * t + 3.33 *t^2 (eq 3) where t = 2 Eq 3 value = 38.32 m < 43 m So accelerating does not work Distance to cover is : 28m + 15 m = 43 m Acceleration is (65-45)/6 = 20/6 = 10.3 = 3.33 m/s^2 Calculate distance for speed + acceleration in 2 secondsin 45*1000/3600 * t + 3.33 *t^2 (eq 3) where t = 2 Eq 3 value = 38.32 m < 43 m So accelerating does not work

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