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) 1 Position, (m)
37.70 9.600
39.30 18.048
40.90 33.792
Calculate the magnitude of the acceleration at t=39.30

Answer 1

The constant acceleration of the object is $0.414\ m/s^2$.

Acceleration of the object

The acceleration of the object is constant, and the values at the three different positions is the same.

For the first position and time

$s = v_0 t + \frac{1}{2} at^2\\\\37.7 = v_o(9.6) +\frac{1}{2} a(9.6)^2\\\\ 37.7 = 9.6v_0 + 46.08a$

For the second position and time

$39.3 = v_o(18.048) +\frac{1}{2} a(18.048)^2\\\\ 39.3 = 18.048v_0 + 162.87a$

Solve the first and second equation together

$37.7 = 9.6v_0 + 46.08a\\\\39.3 = 18.048v_0 + 162.87a\\\\18.048: \ \ 680.41 = 173.261v_0 + 831.651a\\\\9.6: \ \ \ \ -(377.28 = 173.261 v_0+ 1563.552a)\\------------------\\303.13 = -731.901a\\\\-a = \frac{303.13}{731.901} \\\\-a = 0.414 \ m/s^2\\\\|a| = 0.414 \ m/s^2$

Thus, the constant acceleration of the object is $0.414\ m/s^2$.

Learn more about acceleration here: brainly.com/question/14344386