Question

Speedy Sue, driving at 33.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s. Sue applies her brakes but can accelerate only at −1.50 m/s^2 because the road is wet. Will there be a collision? Yes No. 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. (If no, enter "0" for the time.)

Answer 1

Sue's car will collide with the van 726.0 m into the tunnel.

Step-by-step explanation:

When Sue applies her brakes, her acceleration is negative (-1.50 m/s²) due to the road being wet. To determine if there will be a collision, we need to calculate the distance it takes for Sue's car to come to a stop. We can use the equation:

vf² = vi² + 2ax

where vf is the final velocity (0 m/s), vi is the initial velocity (33.0 m/s), a is the acceleration (-1.50 m/s²), and x is the distance. Rearranging the equation to solve for x, we get:

x = (vf² - vi²) / (2a)

Substituting the values, we have:

x = (0² - (33.0 m/s)²) / (2(-1.50 m/s²))

x = 726.0 m

Therefore, Sue's car will collide with the van 726.0 m into the tunnel.