Question

In an experiment, students measure the position x of a cart as a function of time t for a cart that starts at rest and moves with a constant acceleration. The following data are collected for the cart.
t(s) x(m)
0 0
1 4
2 16
3 36
4 64

The acceleration of the cart is most nearly:_____

Answer 1

Given :

Initial velocity , u = 0 m/s² .

To Find :

The acceleration of the cart.

Solution :

Since, acceleration is constant.

Using equation of motion :

$x = ut + \dfrac{at^2}{2}\\\\x = \dfrac{at^2}{2}$

Putting, t = 1 s  and x = 4 m in above equation, we get :

$4 = \dfrac{a(1)^2}{2}\\\\a = 8 \ m/s^2$

Therefore, the acceleration of the cart is 8 m/s².