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
- A \(\bar{v}=\frac{v^2-v_0^2}{a t}\)
- B \(\bar{v}=\frac{v^2+v_0^2}{2 a t}\)
- C \(\bar{v}=\frac{v^2+v_0^2}{a t}\)
- D \(\bar{v}=\frac{v^2-v_0^2}{2 a t}\)
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)]
\(v_{\mathrm{avg}}=\frac{v^2-v_0^2}{2 a t}\)
\(\begin{aligned} & v_1=v_0 \\ & v_2=v\end{aligned}\)
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)]
\(v_{\mathrm{avg}}=\frac{v^2-v_0^2}{2 a t}\)
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