Asked by LilYeet

A runaway truck (truck with no brakes) traveling 45 m/s leaves the highway and enters the runaway ramp where the truck will be slowed due to soft sand that it is made of if the sand slows the truck at -2.00 m/s^2, how long,( distance), must the ramp be in order stop the truck

Answers

Answered by oobleck
v = 45-2t
truck stops at t=22.5
s = 45t - t^2
now evaluate s(22.5)
Answered by R_scott
the stopping time is ... (45 m/s) / (2.00 m/s^2)

the average velocity is ... [(45 m/s) + (0 m/s)] / 2

stopping distance = stopping time * average velocity
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