# The position of a crate sliding down a ramp

The position of a crate sliding down a ramp is given by $x=(0.25t^3)$ m, $y=(1.5t^2)$ m, $z=(6-0.75t^{\frac{5}{2}})$ m, where t is in seconds. Determine the magnitude of the crate’s velocity and acceleration when t = 2 s.

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#### Solution:

Remember that taking the first derivative of a position time equation gives velocity, and take the second derivative of the position time equation gives acceleration. Let us find the velocity and acceleration equations:

$x=(0.25t^3)$ m

$y=(1.5t^2)$ m

$z=(6-0.75t^{\frac{5}{2}})$ m

$v_x=\dot{x}=0.75t^2$ m/s

$v_y=\dot{y}=3t$ m/s

$v_z=\dot{z}=-1.875t^{\frac{3}{2}}$ m/s

$a_x=\ddot{x}=1.5t$ m/$s^2$

$a_y=\ddot{y}=3$ m/$s^2$

$a_z=\ddot{z}=-2.8125t^{\frac{1}{2}}$ m/$s^2$

Remember that a single dot on top represents the first derivative and two dots represent the second derivative. In our case, a single dot represents the velocity, and two dots represent acceleration.

At t = 2 s, the velocity is:

$v_x=0.75(2)^2=3$ m/s

$v_y=3(2)=6$ m/s

$v_z=-1.875(2)^{\frac{3}{2}}=-5.3$ m/s

The magnitude of velocity is:

$v=\sqrt{(v_x)^2+(v_y)^2+(v_z)^2}$

$v=\sqrt{(3)^2+(6)^2+(-5.3)^2}=8.55$ m/s

At t = 2 s, the acceleration is:

$a_x=1.5(2)=3$ m/$s^2$

$a_y=3$ m/$s^2$

$a_z=-2.8125(2)^{\frac{1}{2}}=-3.98$ m/$s^2$

The magnitude of acceleration is:

$a=\sqrt{(a_x)^2+(a_y)^2+(a_z)^2}$

$a=\sqrt{(3)^2+(3)^2+(-3.98)^2}=5.82$ m/$s^2$

$v=8.55$ m/s
$a=5.82$ m/$s^2$