Question #1538

Speedy Sue, driving at 33.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 175 m ahead traveling with velocity 5.10 m/s. Sue applies her brakes but can accelerate only at -2.00 m/s2 because the road is wet. Will there be a collision?If yes, determine how far into the tunnel and at what time the collision occurs. If no, determine the distance of closest approach between Sue's car and the van and enter zero for the time.

Expert's answer

According to the equations of the motion,

h = h_{0} + v_{0}t - at^{2}/2, where h = 175 m, v_{0} = (33.0-5.1) = 27.9 m/s, a = 2 m/s^{2}, h_{0} =0

So we have to solve an equation to find the time:

175 = 27.9t - 2t^{2}/2

t^{2} - 27.9t + 175 =0

D = 8.85^{2}

t_{1} = 18.38 s; t_{2} = 9.52 s.

We have to use the smallest time of 9.52s. At this time the velocity would be V = V_{0}-at = 27.9 - 2*9.52 = 8.85 m/s, it means that the Sue will not have a zero velocity when she reaches the van - and collision will occure at 9.52 seconds after entering the tunnel.

h = h

So we have to solve an equation to find the time:

175 = 27.9t - 2t

t

D = 8.85

t

We have to use the smallest time of 9.52s. At this time the velocity would be V = V

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