Suppose that a senior driving a Pontiac zooms out of a darkened tunnel at 39.4 m/s. She is momentarily blinded by the sunshine. When she recovers, she sees that she is fast overtaking a camper ahead in her lane moving at the slower speed of 12.9 m/s. She hits the brakes as fast as she can (her reaction time is 0.36 s). If she can decelerate at 3.6 m/s^2, what is the minimum distance between the driver and the camper when she first sees it so that they do not collide?

asked by Anonymous

12 years ago

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1 answer

s₀=v₀•t₀=39.4•0.36 =14.2 m
s1=v²-v₀²)/2(-a) = (2.9²-39.4²)/2(-3.6)=5.17 m.
s= s1+s₀=5.17+14.2=19.37 m.

answered by Elena

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Question

Suppose that a senior driving a Pontiac zooms out of a darkened tunnel at 39.4 m/s. She is momentarily blinded by the sunshine. When she recovers, she sees that she is fast overtaking a camper ahead in her lane moving at the slower speed of 12.9 m/s. She hits the brakes as fast as she can (her reaction time is 0.36 s). If she can decelerate at 3.6 m/s^2, what is the minimum distance between the driver and the camper when she first sees it so that they do not collide?

1 answer

s₀=v₀•t₀=39.4•0.36 =14.2 m
s1=v²-v₀²)/2(-a) = (2.9²-39.4²)/2(-3.6)=5.17 m.
s= s1+s₀=5.17+14.2=19.37 m.

Elena · Answer

s₀=v₀•t₀=39.4•0.36 =14.2 m s1=v²-v₀²)/2(-a) = (2.9²-39.4²)/2(-3.6)=5.17 m. s= s1+s₀=5.17+14.2=19.37 m.