# 1019 Velocity and acceleration from the equation of distance | Motion with Variable Acceleration

**Problem 1019**

The motion of a particle is given by the equation $s = 2t^4 - \frac{1}{6}t^3 + 2t^2$ where $s$ is in meter and $t$ in seconds. Compute the values of $v$ and $a$ when $t = 2 \, \text{ sec}$.

**Solution 1019**

## Click here to expand or collapse this section

$s = 2t^4 - \frac{1}{6}t^3 + 2t^2$

$v = 8(2^3) - \frac{1}{2}(2^2) + 4(2)$

$v = \dfrac{ds}{dt}$

$v = 8t^3 - \frac{1}{2}t^2 + 4t$

$a = \dfrac{dv}{dt}$

$a = 24t^2 - t + 4$

When t = 2 sec

$v = 70 \, \text{ m/s}$ *answer*

$a = 24(2^2) - 2 + 4$

$a = 98\, \text{ m/s}^2$ *answer*

- Log in to post comments