# Answer to Question #15259 in Mechanics | Relativity for Amy

Question #15259

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of -5.50 m/s2 for 4.35 s, making straight skid marks 63.0 m long, all the way to the tree. With what speed does the car then strike the tree?

Expert's answer

The car decelerates by

V = -5.50[m/s²]*4.35[s] = -23.925 m/s.

Let's find the initial speed V0 of a car. We know that

V0*4.35[s] - 5.50[m/s²]*(4.35[s])²/2 = 63.0[m],

so&

V0 = (63.0[m] + 5.50[m/s²]*(4.35[s])²/2) / 4.35[s] ≈ 26.445 m/s

and the car strikes the tree with the speed of

V0 + V = 26.445[m/s] - 23.925[m/s] = 2.520 m/s.

V = -5.50[m/s²]*4.35[s] = -23.925 m/s.

Let's find the initial speed V0 of a car. We know that

V0*4.35[s] - 5.50[m/s²]*(4.35[s])²/2 = 63.0[m],

so&

V0 = (63.0[m] + 5.50[m/s²]*(4.35[s])²/2) / 4.35[s] ≈ 26.445 m/s

and the car strikes the tree with the speed of

V0 + V = 26.445[m/s] - 23.925[m/s] = 2.520 m/s.

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