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

Step-by-step 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-(1)/(2)at^(2)`

so

Step 1

a) Let

Step 2

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

I hope this helps you