Answer to Question #234222 in Physics for Jocelyn Lombera

Question

A particle moving along a straight line with constant acceleration has a velocity of 3.56m/s at t= 2.58s and a velocity of -4.83 m/a at t=6.07s what is the particle’a acceleration in m/s^2

Answer 1

By definition, the acceleration is:

$$a = \dfrac{\Delta v}{\Delta t}$$

where $\Delta v$ is the change in velocity that happend in time $\Delta t$. Thus, obtain:

$$a = \dfrac{-4.83m/s - 3.56m/s}{6.07s - 2.58s} \approx -2.40m/s^2$$

Answer. -2.40 m/s^2.