So the stopping distance will be v0trxn +( -v02/2a)
a for this problem is Ff/m = mgμ/m = gμ (note that this acceleration is in the negative direction)
Don't forget to express v0 in m/s
Jhon D.
asked 04/20/22A 2.0 x 103 kg truck is driving down the road at 105 km/hr when a dog runs out on the road 94m in
front of it. The driver’s reaction time is 0.35 seconds before hitting the brakes. The coefficient of
kinetic friction between the time and mud is 0.8. Will the truck hit the dog? (Hint: find the distance
the truck goes before the brakes are put on, then find the distance the truck takes to stop)
So the stopping distance will be v0trxn +( -v02/2a)
a for this problem is Ff/m = mgμ/m = gμ (note that this acceleration is in the negative direction)
Don't forget to express v0 in m/s
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