MCQ
If the velocity of a particle is $v =At+ Bt^2$ ,where $A$ and $B$ are constants, then the distance travelled by it between $1\, s$ and $2\, s$ is
- A$3A+7B$
- ✓$\frac{3}{2}A + \frac{7}{3}B$
- C$\frac{A}{2} + \frac{B}{3}$
- D$\;\frac{3}{2}A + 4B$
✓
Answer
Correct option: B.
$\frac{3}{2}A + \frac{7}{3}B$
b
$V=\alpha t+\beta t^{2}$
$V=\alpha t+\beta t^{2}$
$\frac{ ds }{ dt }=\alpha t +\beta t ^{2}$
$\int_{s_{1}}^{s_{2}} d s=\int_{1}^{2}\left(\alpha t+\beta t^{2}\right) d t$
$S_{2}-S_{1}=\left[\frac{\alpha t^{2}}{2}+\frac{\beta t^{3}}{3}\right]_{1}^{2}$
As particle is not changing direction So distance $=$ displacement.
Distance $=\left[\frac{\alpha[4-1]}{2}+\frac{\beta[8-1]}{3}\right]$
$=\frac{3 \alpha}{2}+\frac{7 \beta}{3}$
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