A sports car moving at constant speed travels 110m in 4.7s a- If it then brakes

Question

A sports car moving at constant speed travels 110m in 4.7s a- If it then brakes and comes to a stop in 4.3s , what is its acceleration in m/s^2 ? b-Express the answer in terms of "g's," where 1.00g=9.8m/s^2.

Explanation / Answer

Firstly, you have to find out the constant speed of the car using the basic equation: S = d / t where S --> constant speed (in m/s) d --> distance traveled (in meter) t --> time taken to travel a given distance (in seconds) Given: d = 110 m t = 4.9 s Compute the speed: S = d / t = 110 m / 4.9 s S = 22.45 m/s Secondly, since you already know the constant speed of the car ( S = 22.45 m/s), you can use this value in computing for the acceleration ( in this case 'deceleration' because the car comes to a stop and therefore the velocity decreases from a certain value to zero). NOTE: The value of speed computed above becomes the value of the starting velocity in computing for acceleration. This is because of the assumption that the sports car is running at that constant speed or velocity before it comes to a full stop. In a straight line motion, speed and velocity do have the same magnitude. From the kinematics equation: vf = vi + a*t where: vf --> final velocity (m/s) vi --> initial velocity/starting velocity (m/s) a --> acceleration (m/s/s) t --> time taken to change the velocity (s) Given: vi = 22.45m/s vf = 0 ; because it comes to a stop t = 4.3 s Find acceleration using the kinematics equation above vf = vi + a*t by rearranging the equation a = (vf -vi) / t a = (0 m/s - 22.45 m/s) / (4.3 s) a = ( -22.45 m/s) / 4.3 s a = - 5.22 m/s/s -->FINAL ANSWER Note: The negative sign of the final answer means the car was decelerating until it comes to a full stop.

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