Coasting due north on your bicycle at 9.5 m/s, you encounter a sandy patch of ro

Question

Coasting due north on your bicycle at 9.5 m/s, you encounter a sandy patch of road 7.8 m across. When you leave the sandy patch your speed has been reduced by 0.8 m/s to 8.7 m/s. (a) Assuming the sand causes a constant acceleration, what was the bicycle's acceleration in the sandy patch? Give both magnitude and direction.

(b) How long did it take to cross the sandy patch?
s

(c) Suppose you enter the sandy patch with a speed of 10.5 m/s. Is your final speed in this case equal to, more than, or less than 9.7m/s?

Explanation / Answer

The patch is 7.8 m wide. Hence the bicycle suffers the decelaration for 7.8 m

We have v^2 = u^2 + 2aS

That 8.7*8.7 = 9.5*9.5 + 2*a*7.8

That is, a = -0.23 m/s^2

The negative sign denotes acceleration opposite to the direction of motion.

b.) Again, v = u + at

hence, t = (v- u)/a = 0.8/0.23 = 3.47 seconds.

c.) For initial velocity to be 10.5, we use the equation v^2 = u^2 + 2as

Hence v^2 = 10.5*10.5 - 2*0.23*7.8 = 10.32 m/s

Therefore, the final velocity in this case will be more than 9.7 m/s

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