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KCET · Physics · Motion In One Dimension

A body is moving along a straight line with initial velocity \(v_0\). Its acceleration \(a\) is constant. After \(t\).seconds, its velocity becomes \(v\). The average velocity of the body over the given time interval is

  1. A \(\bar{v}=\frac{v^2-v_0^2}{a t}\)
  2. B \(\bar{v}=\frac{v^2+v_0^2}{2 a t}\)
  3. C \(\bar{v}=\frac{v^2+v_0^2}{a t}\)
  4. D \(\bar{v}=\frac{v^2-v_0^2}{2 a t}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\bar{v}=\frac{v^2-v_0^2}{2 a t}\)

Step-by-step Solution

Detailed explanation

Given,
\(\begin{aligned} & v_1=v_0 \\ & v_2=v\end{aligned}\)
acceleration \(=a\)
for time \(=t \mathrm{~s}\)
Using Newton's equations of motion,
\(\begin{aligned} v^2-u^2 & =2 a s \\ s & =\frac{v^2-u^2}{2 a}(\text { total distance covered }) \\ s & =\frac{v^2-v_0^2}{2 a}\end{aligned}\)
Average yelocity \(=\frac{\text { Total distance }}{\text { Total time }}=\frac{\frac{v^2-v_0^2}{2 a}}{t}\)
[Using Eq. (i)]
\(v_{\mathrm{avg}}=\frac{v^2-v_0^2}{2 a t}\)