Question

An object moving with uniform acceleration has a velocity of 13.0 cm/s in the positive x-direction when it’s x-coordinate is 2.70 cm. If it’s x-coordinate 2.05 s later is -5.00 cm, what is its acceleration?

Answer 1

8.15 cm / s^2 in the negative x-direction

Explanation

Acceleration is a measure of how fast velocity changes,it is the rate at which velocity changes with time, in terms of both speed and direction. it can be found using the formula.

$d=v_{i}^{\star}t-\frac{1}{2}at^{2}$

so

Step 1

a) Let

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Step 2

$\begin{gathered} d=v_i^*t-\frac{1}{2}at^2 \\ d-v_i^*t=-\frac{1}{2}at^2 \\ 2(d-v_i^*t)=-at^2 \\ \frac{\begin{equation*}2(d-v_i^*t)\end{equation*}}{-t^2}=a \\ a=\frac{2(7.7-13*2.05)}{-(2.05)^2} \\ a=\frac{37.9}{4.2025} \\ a=9.01\text{ }\frac{m}{s^2} \end{gathered}$

I hope this helps you