A driver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of -2.0 m/s^2. How long does it take the car to accelerate to a final speed of 10.0 m/s? How far has the car moved during the braking period?

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A driver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of -2.0 m/s^2. How long does it take the car to accelerate to a final speed of 10.0 m/s? How far has the car moved during the braking period?

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Answer 1 · 2015-07-20T23:17:19

Time needed: 2.5 s
Distance covered: 31.3 m

Explanation:

I'll start with the distance covered while decelerating. Since you know that the initial speed of the car is 15.0 m/s, and that its final speed must by 10.0 m/s, you can use the known acceleration to determine the distance covered by

$v_f^2 = v_i^2 - 2 * a * d$

Isolate $d$ on one side of the equation and solve by plugging your values

$d = (v_i^2 - v_f^2)/(2a)$

$d = ((15.0""^2 - 10.0""^2)"m"^cancel(2)cancel("s"^(-2)))/(2 * 2.0 cancel("m") cancel("s"^(-2))$

$d = color(green)("31.3 m")$

To get the time needed to reach this speed, i.e. 10.0 m/s, you can use the following equation

$v_f = v_i - a * t$ , which will get you

$t = (v_i - v_f)/a$

$t = ((15.0 - 10.0)cancel("m")/cancel("s"))/(2.0cancel("m")/"s"^cancel(2)) = color(green)("2.5 s")$

SIDE NOTE I left the value I got for the distance rounded to three sig figs.

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A driver of a car traveling at 15.0 m/s applies the brakes, causing a uniform acceleration of -2.0 m/s^2. How long does it take the car to accelerate to a final speed of 10.0 m/s? How far has the car moved during the braking period?

1 Answer

Explanation:

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