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?
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 \(2.5\frac{m}{s^{2} }\)