Question

A jet makes a landing traveling due east with a speed of 115 m/s115 \mathrm{~m} / \mathrm{s} If the jet comes to rest in 13.0s13.0 \mathrm{s}, what are the magnitude and direction of its average acceleration?

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Calculate the average acceleration of a jet. The average acceleration formula is given by

aav=ΔvΔt    aav=vfvitfti\begin{align} a_{\text{av}} =\dfrac{\Delta v}{\Delta t} \hspace{5mm} \implies a_{\text{av}} = \dfrac{v_{\text{f}}-v_{\text{i}}}{t_{\text{f}}-t_{\text{i}}} \end{align}

Since the jet brought to rest after 13.0 s, then vf=0v_{\text{f}} = 0. Entering known values to equation(1), we obtain aava_{\text{av}} as

aav=070.6 ms13.0 s=5.43 ms2\begin{align*} a_{\text{av}} &= \dfrac{0-70.6\mathrm{~\dfrac{m}{s}}}{13.0\text{ s}}\\ & = -5.43\mathrm{~\dfrac{m}{s^2}} \end{align*}

The negative value of acceleration denotes the acceleration of the jet is in the opposite direction of its motion. We also choose the positive xx direction to be headed in the east direction. Hence, the magnitude and direction of the acceleration of the jet are then:

aav=5.43 ms2 due west.\boxed{\therefore a_{\text{av}}=5.43\mathrm{~\dfrac{m}{s^2}} \text{ due west.} }

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