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A car low down at -5. 00m/^2 until it come to a top after traveling 15. 0m how much time did it take to top

User Burkestar
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1 Answer

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Answer:

To find the time it took for the car to come to a stop, we need to use the following equation:

v_f^2 = v_0^2 + 2 * a * d

where v_f is the final velocity (0 m/s), v_0 is the initial velocity, a is the acceleration (-5.00 m/s^2), and d is the distance traveled (15.0 m).

Since v_f = 0 m/s, we can simplify the equation:

0 = v_0^2 - 2 * a * d

v_0^2 = 2 * a * d

v_0 = sqrt(2 * a * d)

Next, we can use the value of v_0 to find the time it took to come to a stop:

v_0 = sqrt(2 * a * d)

t = (v_f - v_0) / a

where t is the time it took to come to a stop.

Substituting the values into the equation:

t = (0 - sqrt(2 * (-5.00 m/s^2) * 15.0 m)) / (-5.00 m/s^2)

t = sqrt(30 / 5) / (-5.00)

t = sqrt(6) / (-5.00)

t = 0.7746 / -5.00

t = -0.15493 s

Since time cannot be negative, we will consider the magnitude of the time:

t = abs(-0.15493 s)

t = 0.15493 s

So it took approximately 0.155 seconds for the car to come to a stop.

Step-by-step explanation:

User Daan Van Hulst
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