Question

Speedy Sue, driving at 32.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 170 m ahead traveling with velocity 5.50 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.
Distance in meters?Speed in seconds?

Answer 1

Answer:

10.89 seconds

229.895 m

Explanation:

Distance van travels

$x_v=170+5.5t+\dfrac{1}{2}0t^2\\\Rightarrow x_v=170+5.5t$

Position of car

$x_c=0+32t+\dfrac{1}{2}-2t^2\\\Rightarrow x_c=32t-t^2$

They are equal

$170+5.5t=32t-t^2\\\Rightarrow 17+5.5t-32t+t^2\\\Rightarrow t^2-26.5t+170=0\\\Rightarrow 10t^2-265t+1700=0$

$t=\frac{-\left(-265\right)+\sqrt{\left(-265\right)^2-4\cdot \:10\cdot \:1700}}{2\cdot \:10}, \frac{-\left(-265\right)-\sqrt{\left(-265\right)^2-4\cdot \:10\cdot \:1700}}{2\cdot \:10}\\\Rightarrow t=15.6, 10.89\ s$

The collision occurs at 10.89 seconds

$x_v=170+5.5\times 10.89\\\Rightarrow x_v=229.895\ m$

Collision occurs at 229.895 m from the starting point