rate of acceleration?

Question

A car accelerated from 6.00 m/s to 20.0 m/s, over a distance of 50 m. What is the rate of acceleration??

Answer 1

using eq of motion
v^2 = u^2 + 2as
20^2 = 6^2 + (2 x a x 50)
400 =36 +100a
100a = 364
a = 3.64 ms^-2

Answer 2

Acceleration is defined as the rate of change of velocity, which is given here as 6 m/s and 20 m/s over a distance of 50 m. Using the formula, v^2 - u^2 = 2as, where a is the acceleration and s is the distance, v and u being the final and initial velocities,

20 x 20 - 6 x 6 = 2a x 50 = 100a

400 - 36 = 364 = 100a

a = 364 / 100 = 3.64 m/sec/sec

Answer 3

if the acceleration is constant, then the average velocity is 13m/s (6+20)/2 = 13

50m / 13m/s = 3.85 s

the change in velocity is 14 m/s and the change in time is 3.85 s

therefore 14 m/s / 3.85s = 3.64 m/s^2