## Related questions with answers

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

Solution

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

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

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

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

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

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

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