the velocity of the car has changed from an initial value of 20 m+sec to a final value of 15m+sec in a time interval of 2 sec along positive x direction.what will be the average acceleration during this time travel

Dear Student,

Please find below the solution to the asked query:

Given that the initial velocity of the car is 20 m/s and the final velocity of the car is 15 m/s. The time interval is 2 sec. Therefore, the acceleration acts in the opposite direction to the direction of motion. That is deceleration.

As we know the average of acceleration (deceleration) is the ratio of change in velocity and the time interval.

So, the magnitude of the deceleration is.

$$\begin{gathered} a=\frac{∆v}{∆t}=\frac{v_2-v_1}{∆t}=\frac{15-20}{2}\frac{-5}{2} \\ \Rightarrow a=-2.5 \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right. \end{gathered}$$

Here the negative sign indicates the decrease in velocity.

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