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2.Motion in Straight Line
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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
$3A+7B$
B
$\frac{3}{2}A + \frac{7}{3}B$
C
$\frac{A}{2} + \frac{B}{3}$
D
$\;\frac{3}{2}A + 4B$
(NEET-2016) (JEE MAIN-2021)
Solution
$V=\alpha t+\beta t^{2}$
$\frac{ ds }{ dt }=\alpha t +\beta t ^{2}$
$\int_{s_{1}}^{s_{2}} d s=\int_{1}^{2}\left(\alpha t+\beta t^{2}\right) d t$
$S_{2}-S_{1}=\left[\frac{\alpha t^{2}}{2}+\frac{\beta t^{3}}{3}\right]_{1}^{2}$
As particle is not changing direction So distance $=$ displacement.
Distance $=\left[\frac{\alpha[4-1]}{2}+\frac{\beta[8-1]}{3}\right]$
$=\frac{3 \alpha}{2}+\frac{7 \beta}{3}$
Standard 11
Physics
