21. Antiderivatives, Areas and the FTC

c. From Acceleration to Velocity and Position

On the previous page we discussed \(1\)-dimensional motion with constant acceleration. On this page we discuss \(2\)-dimensional motion with constant acceleration. We use vectors and parametric curves to describe the \(2\)-dimensional motion.

2. Ballistic Motion

When a ball is thrown, it travels horizontally (in the \(x\)-direction) as well as vertically (in the \(y\)-direction). The force of gravity points downward. So the acceleration is the vector: \[ \vec a=\langle0,-g\rangle \]

The initial height is \(y(0)=64\); that's vertical. Since the problem does not specify an initial horizontal position, we take it to be \(x(0)=0\). We put these together as the initial position vector: \[ \vec r(0)=\langle0,64\rangle \]

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In British units, the acceleration is: \[ \vec a=\langle0,-32\rangle \] Using this, we write the velocity as the general antiderivative of the acceleration: (Notice we take the antiderivative of each component.) \[ \vec v=\langle C_1,-32t+C_2\rangle \] We use the initial conditions to find the constants: \[ \vec v(0)=\langle C_1,C_2\rangle=\langle 48\sqrt{3},48\rangle \] So the velocity is: \[ \vec v=\langle 48\sqrt{3},-32t+48\rangle \] Next we write the position as the general antiderivative of the velocity: \[ \vec r=\langle 48\sqrt{3}\,t+K_1,-16t^2+48t+K_2\rangle \] We use the initial conditions to find the constants: \[ \vec r(0)=\langle K_1,K_2\rangle=\langle0,64\rangle \] So the position is: \[ \vec r=\langle 48\sqrt{3}\,t,-16t^2+48t+64\rangle \]

We check by differentiating the position twice and recomputing the initial conditions: \[\begin{aligned} \vec v&=\dfrac{d\vec r}{dt}=\langle 48\sqrt{3},-32t+48\rangle \\ \vec a&=\dfrac{d\vec v}{dt}=\langle0,-32\rangle \end{aligned}\] \[\begin{aligned} \vec r(0)&=\langle 0,64\rangle \\ \vec v(0)&=\langle 48\sqrt{3},48\rangle \end{aligned}\]

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The maximum height is maximum of the \(y\) component of the position. To find the maximum, we set its derivative equal to \(0\), which is setting the \(y\) component of the velocity equal to \(0\): \[ v_y(t)=-32t+48=0 \implies t=\dfrac{48}{32}=\dfrac{3}{2} \] So the maximum height is: \[ y\left(\dfrac{3}{2}\right) =-16\left(\dfrac{3}{2}\right)^2+48\left(\dfrac{3}{2}\right)+64 =-36+72+64\,\text{ft} =100\,\text{ft} \]

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The range is the \(x\) component of the position when the height is \(0\). So we set the \(y\) component of the position equal to \(0\) and solve for the time: \[\begin{aligned} y(t)=-16t^2+48t+64&=0 \\ t^2-3t-4&=0 \\ (t+1)(t-4)&=0 \end{aligned}\] So the height is \(0\) when \(t=-1\) or \(t=4\). Since \(-1\) is in the past, we conclude it hits the ground at \(t=4\). At that time, the horizontal position (the range) is: \[\begin{aligned} x(4)=48\sqrt{3}\,4=192\sqrt{3}\,\text{ft} \end{aligned}\]

The acceleration of gravity on the Moon is about \(\dfrac{1}{6}\) of that on the Earth. We will take it as \[ g=1.6\,\dfrac{\text{m}}{\text{sec}^2} \]

In the problems on this page, the constants of integration all came out to be the values of the initial conditions. This was because the functions were all polynomial and the initial conditions were given at \(t=0\). This will not be the case on the next page.