Answer:
vDP = 21.7454 m/s
θ = 200.3693°
Explanation:
Given
vDE = 7.5 m/s
vPE = 20.2 m/s
Required: vDP
Assume that
vDE to be in direction of - j
vPE to be in direction of i
According to relative motion concept the velocity vDP is given by
vDP = vDE - vPE (I)
Substitute in (I) to get that
vDP = - 7.5 j - 20.2 i
The magnitude of vDP is given by
vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s
θ = Arctan (- 7.5/- 20.2) = 20.3693°
θ is in 3rd quadrant so add 180°
θ = 20.3693° + 180° = 200.3693°
We make use of the equation: v^2=v0^2+2a Δd. We substitute v^2 equals to zero since the final state is halting the truck. Hence we get the equation -<span>v0^2/2a = Δd. F = m a from the second law of motion. Rearranging, a = F/m
</span>F = μ Fn where the force to stop the truck is the force perpendicular or normal force multiplied by the static coefficient of friction. We substitute, -v0^2/2<span>μ Fn/m</span> = Δd. This is equal to
I think its between b or c
