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A roller coaster car has a mass of 290. kilograms. Starting from rest, the car acquires
3.13 × 10^5 joules of kinetic energy as it descends to the bottom of a hill in 5.3 seconds.

Calculate the magnitude of the average acceleration of the roller coaster car as it descends to the bottom
of the hill. [Show all work, including the equation and substitution with units.]

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Kalahira
The answer is 8.8 m/s^2. The acceleration (a) is the change in velocity (v) in time (t): a=Δv/t. Δv=v2-v1. Since the car starts from the rest, v1=0 m/s. The final velocity (v2) can be calculated from kinetic energy (KE) and the mass (m) of the car since KE=1/2*m*v2^2. From here: v2^2=2KE/m. v2=√(2KE/m). KE=3.13×10^5 J=3.13×10^5 kg*m^2/s^2, m=290 kg. v2=√(2*3.13×10^5 kg*m^2/s^2/290 kg)=√(2,158.62 m^2/s^2)=46.5 m/s. So, Δv=v2-v1=46.5 m/s - 0 m/s= 46.5 m/s. Now, we have Δv = 46.5 m/s and t=5.3 s, so the acceleration is: a=Δv/t=46.5 m/s/5.3 s = 8.8 m/s^2.
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