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Mnenie [13.5K]
3 years ago
6

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
Physics
2 answers:
schepotkina [342]3 years ago
8 0

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.

emmasim [6.3K]3 years ago
5 0

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.

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