The acceleration of a particle increases linearly with time t as 6 t . If the initial velocity of the particle is zero and the particle starts from the origin, then the distance travelled by the particle in time t will be

Question

The acceleration of a particle increases linearly with time t as 6 t . If the initial velocity of the particle is zero and the particle starts from the origin, then the distance travelled by the particle in time t will be (A) t (B) t2 (C) t3 (D) t4

Tardigrade Admin · Accepted Answer

We know that a=(d v/d t) Then according to the question (d v/d t)=6t or dv=6tdt,v=(6 t2/2)=3t2 , Also, dx=3t2dt ⇒ x=3(t3/3)=t3

Q. The acceleration of a particle increases linearly with time $t$ as 6 $t$ . If the initial velocity of the particle is zero and the particle starts from the origin, then the distance travelled by the particle in time $t$ will be

$t$

$t^{2}$

$t^{3}$

$t^{4}$

We know that $a = \frac{d v}{d t}$
Then according to the question
$\frac{d v}{d t} = 6 t$ or $d v = 6 t d t, v = \frac{6 t ^{2}}{2} = 3 t^{2}$ ,
$A l so, d x = 3 t^{2} d t \Rightarrow x = 3 \frac{t ^{3}}{3} = t^{3}$

Q. The acceleration of a particle increases linearly with time t as 6 t . If the initial velocity of the particle is zero and the particle starts from the origin, then the distance travelled by the particle in time t will be

t

t2

t3

t4

We know that a=dtdv​ Then according to the question dtdv​=6t or dv=6tdt,v=26t2​=3t2 , Also,dx=3t2dt⇒x=33t3​=t3

Q. The acceleration of a particle increases linearly with time $t$ as 6 $t$ . If the initial velocity of the particle is zero and the particle starts from the origin, then the distance travelled by the particle in time $t$ will be

$t$

$t^{2}$

$t^{3}$

$t^{4}$

We know that $a = \frac{d v}{d t}$
Then according to the question
$\frac{d v}{d t} = 6 t$ or $d v = 6 t d t, v = \frac{6 t ^{2}}{2} = 3 t^{2}$ ,
$A l so, d x = 3 t^{2} d t \Rightarrow x = 3 \frac{t ^{3}}{3} = t^{3}$