Question

.A cart rolling down an incline for 5.0 seconds has an acceleration of 1.6 m/s2. If the cart has a beginning speed of 2.0 m/s, and its final velocity of 10 m/s, what was the car's displacement?

Answer 1

Use the formula,

$\Delta x=v_it+\dfrac12at^2$

where $\Delta x$ is the cart's displacement (from the origin), $v_i$ is its initial speed, $a$ is its acceleration, and $t$ is time.

$\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2$

$\implies\boxed{\Delta x=30\,\mathrm m}$

Alternatively, since acceleration is constant, we have

$\dfrac{v_f+v_i}2=\dfrac{\Delta x}t$

That is, we have these two equivalent expressions for average velocity, where $v_f$ is the cart's final velocity. Solve for $\Delta x$:

$\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}$

$\implies\boxed{\Delta x=30\,\mathrm m}$