Jim is driving a 2268-kg pickup truck at 30.0 m/s and releases his foot from the

Question

Jim is driving a 2268-kg pickup truck at 30.0 m/s and releases his foot from the accelerator pedal. The car eventually stops due to an effective friction force that the road, air, and other things exert on the car. The friction force has an average magnitude of 900 N .

Determine the average acceleration while stopping of the truck.

Express your answer with the appropriate units.

Would a heavier car travel farther before stopping or stop sooner if the average friction force were the same?

the stopping distance will be longer the stopping distance will be shorter the stopping distance will be the same

Explanation / Answer

Avg acceleraation =averaage friction force/ mass of truck

= 900/2268

=0.39 m/s2

The heavier car will have higher momentum ,and hence "the stopping distance will be longer".

Reason:

Force = rate of change of momentum.

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