A particle is moving in a straight line. Its velocity (v) obeys a relation with time ( t) given by v=(1/3)(3 t2+2 t) m / s. Find the distance travelled (in m ) by the particle in a time interval of t=2 s to t=3 s.

Question

A particle is moving in a straight line. Its velocity (v) obeys a relation with time ( t) given by v=(1/3)(3 t2+2 t) m / s. Find the distance travelled (in m ) by the particle in a time interval of t=2 s to t=3 s. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

v =(1/3)(3 t2+2 t) or (d s/d t)=(1/3)(3 t2+2 t) ⇒ s=∫ d s=(1/3)=[∫ limits23(3 t2+2 t) d t] =(1/3)[(t3+t2)]23 =(1/3)[(33-23)+(32-22)] =(1/3)[19+5]=8 m

Q. A particle is moving in a straight line. Its velocity $(v)$ obeys a relation with time ( $t)$ given by $v = \frac{1}{3} (3 t^{2} + 2 t) m / s$ . Find the distance travelled (in $m$ ) by the particle in a time interval of $t = 2 s$ to $t = 3 s$ .

$v = \frac{1}{3} (3 t^{2} + 2 t)$
or $\frac{d s}{d t} = \frac{1}{3} (3 t^{2} + 2 t)$
$\Rightarrow s = \int d s = \frac{1}{3} = [2 \int 3 (3 t^{2} + 2 t) d t]$
$= \frac{1}{3} [(t^{3} + t^{2})]_{2}^{3}$
$= \frac{1}{3} [(3^{3} - 2^{3}) + (3^{2} - 2^{2})]$
$= \frac{1}{3} [19 + 5] = 8 m$

Q. A particle is moving in a straight line. Its velocity (v) obeys a relation with time ( t) given by v=31​(3t2+2t)m/s. Find the distance travelled (in m ) by the particle in a time interval of t=2s to t=3s.

v=31​(3t2+2t) or dtds​=31​(3t2+2t) ⇒s=∫ds=31​=[2∫3​(3t2+2t)dt] =31​[(t3+t2)]23​ =31​[(33−23)+(32−22)] =31​[19+5]=8m

Q. A particle is moving in a straight line. Its velocity $(v)$ obeys a relation with time ( $t)$ given by $v = \frac{1}{3} (3 t^{2} + 2 t) m / s$ . Find the distance travelled (in $m$ ) by the particle in a time interval of $t = 2 s$ to $t = 3 s$ .

$v = \frac{1}{3} (3 t^{2} + 2 t)$
or $\frac{d s}{d t} = \frac{1}{3} (3 t^{2} + 2 t)$
$\Rightarrow s = \int d s = \frac{1}{3} = [2 \int 3 (3 t^{2} + 2 t) d t]$
$= \frac{1}{3} [(t^{3} + t^{2})]_{2}^{3}$
$= \frac{1}{3} [(3^{3} - 2^{3}) + (3^{2} - 2^{2})]$
$= \frac{1}{3} [19 + 5] = 8 m$