Question

Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.40 m/s. Sue applies her brakes but can accelerate only at −1.90 m/s2 because the road is wet. Will there be a collision? Yes No

Answer 1

Answer:

Yes there will be a collision.

Explanation:

We shall use the relative motion between the 2 cars to solve

Relative speed of Sue with respect to slow moving van is

$V_{r}=V_{sue}-V_{van}\\\\V_{r}=34-5.40=28.6m/s$

Now in order to stop this vehicle the distance it requires to stop shall be calculated by third equation of kinematics as

$v^{2}=u^{2}+2as$

since the finally the car shall stop thus v = 0 m/s

Applying values in the above equation and solving for 's' we get

$0=28.6^{2}+2\times -1.90\times s\\\\\therefore s=\frac{28.6^{2}}{2\times 1.90}\\\\s=215.25m$

Since distance required to stop is greater than available distance of 160 m thus the two cars shall collide.

Answer 2

Answer:

Yes, there will be collision.

Explanation

Initial relative velocity u  = 34-5.4 = 28.6 m/s

required final relative velocity  v = 0

Acceleration , suppose be a .

Distance traveled =d = 160 m

Using the formula for relative motion,

v² - u² = 2as

0 - 28.6²=2as

a = - 2.55 m s⁻².

Since this required value is more than maximum  braking deceleration so , there will be collision.