Question

An object is moving in a straight line with a constant acceleration. Its position is measured at three different times, as shown in the table below. Time (s) Position, (m) 52.50 8.900 53.90 15.704 55.30 26.036 Calculate the magnitude of the acceleration at t-53.90s.

Answer 1

Here , for the object between 1 and second position

displacement = 15.704 - 8.9 = 6.804 m

time taken = 53.90 - 52.50 = 1.4 s

Now, let the constant acceleration is a

and the initial velocity at the first location is u

Using second equation of motion

displacement = u * t + 0.50at^2

6.804 = u * 1.4 + 0.50 * a * 1.4^2 -----(1)

Now, between position 1 and 3

displacement = 26.036 - 8.9 = 17.136 m

time taken = 55.3 - 52.50 = 2.8 s

and the initial velocity at the first location is u

Using second equation of motion

displacement = u * t + 0.50at^2

17.136 = u * 2.8+ 0.50 * a * 2.8^2 ----(2)

solving equations 1 and 2

u = 3.6 m/s

a = 1.8 m/s^2

the constant acceleration is 1.8 m/s^2 at all times