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a car in a drag race started from rest and accelerated constantly to a velocity of 50 m/s when it reached the end of a 500 m road. What was the cars rate of acceleration?

Verified Answer :

Use the kinematic equation, \(v _{f} ^{2} = v _{o} ^{2} + 2a ( x_{f} - x_{o} )\) 
use your givens, v initial is 50m/s, v final would be 0m/s since the car is coming to a rest, x final would be 500m, x intial would be 0m, and you are looking for your acceleration. plug in your givens into your equation:
 \((50) ^{2} = (0) ^{2} + 2(a)(500)\)

so, by solving for a, you would get 2.5m/s\(^{2}\)

Answer:

The acceleration of the car is \(2.5\frac{m}{s^{2} }\)

Explanation:

Kinematics of the problem:

The car moves with uniform accelerated movement, then, We use the acceleration equation for this type of movement:

\(a=\frac{1}{2*d} (v_{f} ^{2} -v_{i} ^{2} )\)  Formula 1

vf: final speed=\(50\frac{m}{s}\)

vi: initial speed=0

d=distance traveled by the car=500m


Acceleration calculating:


We replace known information in the equation 1:

\(a=\frac{50^{2} -0}{2*500}\)

\(a=\frac{2500}{1000}\)

\(a=2.5\frac{m}{s^{2} }\)

Answer:

The acceleration of the car is \(2.5\frac{m}{s^{2} }\)