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EXAMPLE 7Analyzing the Acceleration of a Car

A car on the ramp of a multistory parking garage travels along a curve traced out by r(t)=10costi+10sintj+3tk, where t is the time in hours and distance is in miles. Find the tangential component aT and normal component aN of the acceleration of the car. What are the magnitude and direction of the force on the driver?

Solution We begin by finding the velocity v, speed v, and acceleration a of the car. v(t)=r′(t)=ddt(10cost)i+ddt(10sint)j+ddt(3t)k=−10sinti+10costj+3kv(t)=‖

Since the speed is constant, the tangential component of acceleration a_{ \mathbf{T}} is \begin{eqnarray*} a_{\mathbf{T}}=\dfrac{dv}{dt}=0 \end{eqnarray*}

The normal component of acceleration a_{\mathbf{N}} is

Since, \begin{equation*} \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] -10\sin t & 10\cos t & 3 \\[3pt] -10\cos t & -10\sin t & 0 \end{array}\right| =30\sin t\mathbf{i}-30\cos t\mathbf{j}+100\mathbf{k} \end{equation*}

the normal component is \begin{eqnarray*} a_{\mathbf{N}}&=&\frac{\left\Vert 30\sin t\mathbf{i}-30\cos t\mathbf{j}+100 \mathbf{k}\right\Vert }{\sqrt{109}}=\dfrac{\sqrt{900\sin ^{2}t+900\cos ^{2}t+10000} }{\sqrt{109}}\\[4pt] &=&\dfrac{\sqrt{10900}}{\sqrt{109}}=10 \end{eqnarray*}

As the car travels on the ramp at a constant speed, the force \mathbf{F}=m\mathbf{a}\;=\;ma_{\mathbf{N}} \mathbf{N}\;=\;10 m \mathbf{N}

pulls the car and driver toward the center of the ramp, with a magnitude about 10 times the mass of the car.