The velocity \(v_x\) of a particle moving along the \(x\) - axis varies according to the expression \(v_x=(10\) \(\left.-5 t^2\right) \mathrm{ms}^{-1}\), where \(\mathrm{t}\) is in seconds. Calculate the average acceleration in the time interval \(\mathrm{t}=0\) to \(t=-2\) seconds
A. \(-5 \mathrm{~ms}^{-2}\) B. \(-15 \mathrm{~ms}^{-2}\) C. \(-20 \mathrm{~ms}^{-2}\) D. \(-10 \mathrm{~ms}^{-2}\)
Correct Answer: C
Explanation
given: \(v_x=10-5 t^2\) required average acceleration \(=\) ? average acceleration \(=\frac{\text { changc in vclocity }}{\text { change in time }}\) when \(\mathrm{t}=0 \mathrm{~s}\) \(\mathrm{v}_0=10-5(0)^2=10 \mathrm{~m} / \mathrm{s}\) when \(\mathrm{t}=-2 \mathrm{~s}\) \(\mathrm{v} \cdot-2=10-5(-2)^2\) \(=10-5(4)=10-20=-10 \mathrm{~m} / \mathrm{s}\) \(\therefore \mathrm{a}=\frac{\mathrm{v}_0-\mathrm{v}_{-2}}{\mathrm{t}_0-\mathrm{t}_{-2}}\) \(=\frac{10-(-10)}{0-(-2)}=\frac{20}{2}=10 \mathrm{~m} / \mathrm{s}^2\)
