A person driving her car at 49 km/h approaches an intersection just as the traff

Question

A person driving her car at 49 km/h approaches an intersection just as the traffic light turns yellow. The yellow light lasts only 2.0 s before turning to red, and she is 30 m away from the near side of the intersection (Fig. 2-29). The intersection is 15 m wide. Her car's maximum deceleration is -5.4 m/s2, whereas it can accelerate from 49 km/h to 70 km/h in 6.3 s. Ignore the length of her car and her reaction time. Figure 2-29 If she hits the brakes, how far will she travel before stopping? m If she hits the gas instead, how far will she travel before the light turns red? m Should she try to stop, or should she speed up to cross the intersection before the light turns red?

Explanation / Answer

First, you have to convert 49 km/h into m/s 49 km/h = 49000 m/h = 13.61 m/s The formula to find the time it would take her to stop is v = vi + a * delta(t) 0 = 13.61 m/s + (-5.4 m/s^2) * delta (t) delta(t) = 2.52 seconds Now that we know how long it's going to take her to stop, we can calculate how far she will travel: s = si + 1/2 * (v + vi) * delta(t) s = 0 + 1/2 * (0 m/s + 13.61 m/s) * 2.52 s s = 17.154 meters She will travel 17.154 meters before comming to a stop. For the second question, we need to calculate the acceleration of the car. Since the acceleration: a = v - vi / t a = 70 km/h - 49 km/h / 6.3 s a = 21 km/h / 6.3 s a = 5.83 m/s / 6.3 s a = 0.925 m/s^2 We then want to know the distance traveled in 2 seconds (the time before the light turn red) d = vi * t + a * t^2 d = 49 km/h * 2 s + 0.925 m/s^2 * (2 s)^2 d = 13.61 m/s * 2 s + 0.925 m/s^2 * 4 s^2 d = 27.22 m + 3.703 m d = 30.923 m Before the light turns red, she will have traveled 30.923 m. Therefore, she will be at the very beginning of the intersection before the light turns red.

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