# 2-47 - The driver has a speed v0 and is at position x=0...

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The driver has a speed v 0 and is at position x =0 when he sees the deer at x =35. After 0.5 s his position is 0 0 5 . 0 v x = . So, now, the equations of motion are t v v t t v v x 10 5 5 . 0 0 2 0 0 - = - + = . We are told that when , 0 = v 35 = x . So, t v 10 0 = and therefore 2 2 2 5 5 5 10 5 35 t t t t t + = - + = and so - = 2 . 3 2 . 2 t seconds (solving the quadratic equation). Now, there are two solutions which will have the car stop at 35 = x m, - = 32 22 0 v m/s. We know that 0 0 < v cannot be true, so the positive time solution must be the right one. The correct equations of motion for the car thus become t v t t x 10 22 5 22 11 2 - = - + = . For these equations of motion, there is only one time when 0 = v because a parabola has only one extremum. The graph shows the car starting at 11 = x m and stopping at 35 = x m at 2 . 2 = t s. 0 0.5 1 1.5 2 2.5 t 0 10 20 30 40 x

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Just for the fun of it, let’s examine the other solution, 2 . 3 -
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## 2-47 - The driver has a speed v0 and is at position x=0...

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