CALCULATING AVERAGE AND INSTANTANEOUSE ACCLERATION
Let’s return to the motions of the Mars rover in Example 3.1.
(a) Find the components of the average acceleration for the interval t = 0.0 s to t = 2.0 s.
(b) Find the instantaneous acceleration at t = 2.0 s.
IDENTIFY and SET UP:
In Example 3.1 we found the components of the rover’s instantaneous velocity at any time t:
v_{x}=\frac{d x}{d t}=\left(-0.25 \mathrm{~m} / \mathrm{s}^{2}\right)(2 t)=\left(-0.50 \mathrm{~m} / \mathrm{s}^{2}\right) t
v_{y}=\frac{d y}{d t}=1.0 \mathrm{~m} / \mathrm{s}+\left(0.025 \mathrm{~m} / \mathrm{s}^{3}\right)\left(3 t^{2}\right)=1.0 \mathrm{~m} / \mathrm{s}+\left(0.075 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}
We’ll use the vector relationships among velocity, average acceleration, and instantaneous acceleration. In part (a) we determine the values of v_x and v_y at the beginning and end of the interval and then use Eq. (3.8)(\vec{a}_{\mathrm{av}}=\frac{\Delta \overrightarrow{\boldsymbol{v}}}{\Delta t}=\frac{\overrightarrow{\boldsymbol{v}}_{2}-\overrightarrow{\boldsymbol{v}}_{1}}{t_{2}-t_{1}}) to calculate the components of the average acceleration. In part (b) we obtain expressions for the instantaneous acceleration components at any time t by taking the time derivatives of the velocity components as in Eqs. (3.10) (a_{x}=\frac{d v_{x}}{d t} \quad a_{y}=\frac{d v_{y}}{d t} \quad a_{z}=\frac{d v_{z}}{d t}).
EXECUTE:
(a) In Example 3.1 we found that at t = 0.0 s the velocity components are
v_{x}=0.0 \mathrm{~m} / \mathrm{s} \quad v_{y}=1.0 \mathrm{~m} / \mathrm{s}and that at t = 2.0 s the components are
v_{x}=-1.0 \mathrm{~m} / \mathrm{s} \quad v_{y}=1.3 \mathrm{~m} / \mathrm{s}Thus the components of average acceleration in the interval t = 0.0 s to t = 2.0 s are
a_{\mathrm{av}-x}=\frac{\Delta v_{x}}{\Delta t}=\frac{-1.0 \mathrm{~m} / \mathrm{s}-0.0 \mathrm{~m} / \mathrm{s}}{2.0 \mathrm{~s}-0.0 \mathrm{~s}}=-0.50 \mathrm{~m} / \mathrm{s}^{2}a_{\mathrm{av}-y}=\frac{\Delta v_{y}}{\Delta t}=\frac{1.3 \mathrm{~m} / \mathrm{s}-1.0 \mathrm{~m} / \mathrm{s}}{2.0 \mathrm{~s}-0.0 \mathrm{~s}}=0.15 \mathrm{~m} / \mathrm{s}^{2}
(b) Using Eqs. (3.10) (a_{x}=\frac{d v_{x}}{d t} \quad a_{y}=\frac{d v_{y}}{d t} \quad a_{z}=\frac{d v_{z}}{d t}), we find
a_{x}=\frac{d v_{x}}{d t}=-0.50 \mathrm{~m} / \mathrm{s}^{2} \quad a_{y}=\frac{d v_{y}}{d t}=\left(0.075 \mathrm{~m} / \mathrm{s}^{3}\right)(2 t)Hence the instantaneous acceleration vector \overrightarrow{\boldsymbol{a}} at time t is
\vec{a}=a_{x} \hat{\imath}+a_{y} \hat{\jmath}=\left(-0.50 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\imath}+\left(0.15 \mathrm{~m} / \mathrm{s}^{3}\right) t \hat{\jmath}At t = 2.0 s the components of acceleration and the acceleration vector are
a_{x}=-0.50 \mathrm{~m} / \mathrm{s}^{2} \quad a_{y}=\left(0.15 \mathrm{~m} / \mathrm{s}^{3}\right)(2.0 \mathrm{~s})=0.30 \mathrm{~m} / \mathrm{s}^{2} \vec{a}=\left(-0.50 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\imath}+\left(0.30 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\jmath}The magnitude of acceleration at this time is
a=\sqrt{a_{x}^{2}+a_{y}^{2}}=\sqrt{\left(-0.50 \mathrm{~m} / \mathrm{s}^{2}\right)^{2}+\left(0.30 \mathrm{~m} / \mathrm{s}^{2}\right)^{2}}=0.58 \mathrm{~m} / \mathrm{s}^{2}
A sketch of this vector (Fig. 3.9) shows that the direction angle β of \overrightarrow{\boldsymbol{a}} with respect to the positive x-axis is between 90° and 180°.
From Eq. (3.7) (\tan \alpha=\frac{v_{y}}{v_{x}}) we have
Hence β = 180° + 1-31°2 = 149°.
EVALUATE: Figure 3.9 shows the rover’s path and the velocity and acceleration vectors at t = 0.0 s, 1.0 s, and 2.0 s. (Use the results of part (b) to calculate the instantaneous acceleration at t = 0.0 s and t = 1.0 s for yourself.) Note that \overrightarrow{\boldsymbol{v}} \text { and } \overrightarrow{\boldsymbol{a}} are not in the same direction at any of these times. The velocity vector \overrightarrow{\boldsymbol{v}} is tangent to the path at each point (as is always the case), and the acceleration vector \overrightarrow{\boldsymbol{a}} points toward the concave side of the path.