The instantaneous velocity


The instantaneous velocity of a particle moving in a straight line is given as $\mathrm{v}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}$, where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between $1 \mathrm{~s}$ and $2 \mathrm{~s}$ is :

  1. $3 \alpha+7 \beta$

  2. $\frac{3}{2} \alpha+\frac{7}{3} \beta$

  3. $\frac{\alpha}{2}+\frac{\beta}{3}$

  4. $\frac{3}{2} \alpha+\frac{7}{2} \beta$

Correct Option: , 2


$\mathrm{V}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}$

$\frac{\mathrm{ds}}{\mathrm{dt}}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}$

$\int_{s_{1}}^{s_{2}} d s=\int_{1}^{2}\left(\alpha t+\beta t^{2}\right) d t$

$\mathrm{S}_{2}-\mathrm{S}_{1}=\left[\frac{\alpha \mathrm{t}^{2}}{2}+\frac{\beta \mathrm{t}^{3}}{3}\right]_{1}^{2}$

As particle is not changing direction

So distance $=$ displacement.

Distance $=\left[\frac{\alpha[4-1]}{2}+\frac{\beta[8-1]}{3}\right]$

$=\frac{3 \alpha}{2}+\frac{7 \beta}{3}$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now