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.

The position of a crate sliding down a ramp

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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{(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{(3)^2+(3)^2+(-3.98)^2}=5.82 m/s^2


Final Answers:

v=8.55 m/s

a=5.82 m/s^2


This question can be found in Engineering Mechanics: Dynamics (SI edition), 13th edition, chapter 12, question 12-81.

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