Finding the Velocity, Acceleration, and Speed of a Vector Particle in 3D (Example 3)

Here's an example where you're given the acceleration  function, and need to integrate to get velocity and position functions.

A ball is thrown from the origin (0,0,0). The initial velocity is \(50i + 80k\) in feet/seconds and the acceleration of the ball is given by the vector function \(a(t) = - 4j - 32k\). What time does it land at and at what velocity and also what speed does the ball hit the ground?

The first step is to get the velocity function. Let's rewrite in vector bracket notation:

\[a(t) = \left\langle {0, - 4, - 32} \right\rangle \]

Note that when we integrate we do it with respect to time, "t". Acceleration is just made of a vector of constants, so we just add "t". Note that we'll get a constant from integrating, since all of the components have the same constant we can take it out of the vector brackets.

\[v(t) = \left\langle {0, - 4t, - 32t} \right\rangle + {v_O}\]

This constant of integration can be solved for by using the initial velocity condition. The initial velocity was provided in the question and is:

\[v(0) = \left\langle {50,0,80} \right\rangle \]

So substitute that in and we'll get v(t):

\[v(t) = \left\langle {50, - 4t,80 - 32t} \right\rangle \]

Integrate this again to get the position vector function r(t).

\[r(t) = \left\langle {50t, - 2{t^2},80t - 16{t^2}} \right\rangle + {r_O}\]

We get another constant that I factored out. In the question we're told the ball is thrown from the origin, (0,0,0), so

\[r(0) = \left\langle {0,0,0} \right\rangle \]


\[r(t) = \left\langle {50t, - 2{t^2},80t - 16{t^2}} \right\rangle \]

So that's our position vector function above. When does the ball hit the ground? To find that out, set the final component, the "z" or "k" component to 0, the other components don't matter at the moment.

I've been asked why to use the last component as height, it's because it is a universal notation to use the third component as the "z" component or height.

Setting the third component/dimension to be 0,

\[0 = 80t - 16{t^2}\]

Solving the quadratic, t = 0 or t = 5. The ball started at the origin, so when it lands again it will be at 5 seconds.

We plug t=5 into v(t) to get the velocity vector it hits the ground at, then we can find the norm of that to get the speed at which it hits the ground.

\[v(5) = \left\langle {50, - 4(5),80 - 32(5)} \right\rangle \]

\[v(5) = \left\langle {50, - 20, - 80} \right\rangle \]

v(5) is the velocity it hits the ground at, in units of feet/second since that was the units told in the question. And now to find the speed, it's just the norm of v(5)'s vector.

\[\left| {v(5)} \right| = \sqrt {{{50}^2} + {{( - 20)}^2} + {{( - 80)}^2}} \]

\[\left| {v(5)} \right| = \sqrt {9300} \]

\[\left| {v(5)} \right| = \sqrt {(100)(93)} \]

Remember that if you can a perfect squared value (4, 9, 16, 25, 36, etc.) under a root you can square root it and take it out of the root.

\[\left| {v(5)} \right| = 10\sqrt {93} \]

And the speed will also be in units of feet/sec.

Please see our other worked-out examples on finding displacement, velocity, speed, and acceleration of a moving particle:

Velocity Acceleration Speed of Vector Particle 3D (Example 1)</p> <p>Alternatively return to the particle in motion hub page:</p> <p>

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