Speedy Sue, driving at 30.0 m/s, enters a one lane tunnel at time t=0. At that m

Question

Speedy Sue, driving at 30.0 m/s, enters a one lane tunnel at time t=0. At that moment she observes a van 155 m ahead, traveling in the same direction at a constant speed of 5.00 m/s. Sue immediately applies her brakes but can accelerate only at -2.00 m/s^2 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 closest approach between Sue's car and the van.

I know it is Yes, there will be a collision but I am not sure how far into the tunnel and the time the collision occurs. I am stuck here.

Explanation / Answer

To solve this, you should find the time it takes for sue to reach the speed of the van, and where she will be:

Vf = Vi + a*t
5 = 30 + (-2) * t
t = 12.5 seconds

To find the distance she travels:

Vf^2 = Vi^2 + 2 * a * d
5^2 = 30^2 + 2*(-2)*d
d = 218.75 meters

To find the distance the van travels:

d = V*t
d = 5 m/s * 12.5 s
d = 62.5 meters

Since the van started 155 meters ahead of sue, it would be at a distance of 155 m + 62.5 m = 217.5 m from her initial position. So there will be a collision.

To get how far into the tunnel, you need the time when their distances are the same:

For the van:

d = 155 + V*t

d = 155 + 5*t

For the car, you have:

d = V*t + (1/2)*a*t^2

d = 30*t - t^2

Since you can set the two equations equal to each other (both equal the same d), you get:

155 + 5*t = 30*t - t^2

using the quadratic equation: t = 11.38 seconds

To get the distance into the tunnel the collide, you can then use:

d = 30*t - t^2

d = 30*11.38 - 11.38^2

d = 211.90 meters into the tunnel

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.