Please help with magnitude of acceleration problem...

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)
50.10s, 8.800 m
51.40s, 14.507 m
52.70s, 24.608 m

Calculate the magnitude of the acceleration at t=51.40s.

To get the acceleration, you need the velocity at 2 different times. You are given the positions at three different times, times 1, 2 and 3. You can get the average velocity between times 1 and 2 using v_avg = delta_x/delta_t. This average velocity is the same as the instantaneous velocity at the time midway between times 1 and 2. Do the same for times 2 and 3.

so it should be around 2.6 m/s^2

The magnitude of acceleration does not change. V = (14.507-8.8) / (51.4-50.10)

a = V /(51.4-50.10) = 3.377 m/s^2