Question

A particle has an initial horizontal velocity of 1.9 m/s and an initial upward velocity of 3.7 m/s. It is then given a horizontal acceleration of 1.1 m/s 2 and a downward acceleration of 1.2 m/s 2 . What is its speed after 3.4 s? Answer in units of m/s.

Answer 1

Its horizontal and vertical velocities are given respectively by

$v_x=1.9\dfrac{\rm m}{\rm s}+\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)t$

$v_y=3.7\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)t$

After $t=3.4\,\mathrm s$, the components of its velocity are

$v_x=1.9\dfrac{\rm m}{\rm s}+\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(3.4\,\mathrm s)=5.64\dfrac{\rm m}{\rm s}$

$v_y=3.7\dfrac{\rm m}{\rm s}+\left(-1.2\dfrac{\rm m}{\mathrm s^2}\right)(3.4\,\mathrm s)=-0.38\dfrac{\rm m}{\rm s}$

Its overall speed is the magnitude of its velocity:

$\|\vec v\|=\sqrt{{v_x}^2+{v_y}^2}\approx5.7\dfrac{\rm m}{\rm s}$