30 km/h * 1000 m/km * 1 h/3600 s = 8.33 m/s
50 * 1000/3600 = 13.9 m/s
v = Vi + a t
8.33 = 13.9 + at
so
a t = - 5.56 m/s^2
so
t = -5.56/a
d = Vi t + (1/2) a t^2
150 = 13.9 (-5.56/a) + .5 a (-5.56/a)^2
150 = -77.3/a + 15.5/a
150 a = - 61.8
a = -.412
magnitude of a = .412
Approaching a flashing pedestrian activated traffic light, a driver must slow down to a speed of 30km/h. If the cross walk is 150m away and the vehicle's initial speed us 59km/h, what must be the magnitude of the car's acceleration to reach the speed limit?
Answer is 0.41m/s^2(up)
1 answer