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krok68 [10]
3 years ago
15

A car traveling at 40 ft/sec decelerates at a constant 5 feet per second per second. how many feet does the car travel before co

ming to a complete stop?
Physics
1 answer:
Tema [17]3 years ago
3 0
<span>This problem is solved by the equation of motion: x = x0 + v0*t + 1/2*a*t^2, Here x0 = 0, v0 = 40ft/sec and a = -5 ft/s^2, we need to solve for t: v = v0 + a*t, solve how long does it take to stop: 0 = v0 + a*t --> a*t = -v0 --> t = -v0/a -- > 40/5 = 8 seconds to stop. In this time, the car travels x = 0 + 40*8 + 0.5*-5*8^2 ft ~ 160 ft. Answer: The car travels 160 ft.</span>
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a.Let the sidewalk moving in positive x- direction.

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Time=\frac{distance}{speed}

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving=\frac{35}{v_w}=\frac{35}{2.5}=14s

b.If she gets on at the end opposite the end in part (a)

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Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground=v_w=v_s+v=1-1.5=-0.5m/s

Time=\frac{-35}{-0.5}=70 s

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

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