The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with an initial velocity v0. The distance travelled by the particle in time t will be

Question

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with an initial velocity v0. The distance travelled by the particle in time t will be (A) v0t+(1/3)bt2 (B) v0 t+(1/2)bt2 (C) v0 t+(1/6)bt3 (D) v0 t+(1/3)bt3

Tardigrade Admin · Accepted Answer

Acceleration propto bt. i.e., (d2 x/dt2) = a propto bt Integrating, (dx/dt) =(bt2/2)+C Initially, t = 0, dx/dt=v0 Therefore, (dx/dt) =(bt2/2)+v0 Integrating again, x = (bt3/6)+v0 t + C When t = 0, x = 0 ⇒ C = 0. i.e., distance travelled by the particle in time t = v0 t + (bt3/6).

Q. The acceleration of a particle is increasing linearly with time $t$ as $b t$ . The particle starts from origin with an initial velocity $v_{0}$ . The distance travelled by the particle in time t will be

$v_{0} t + \frac{1}{3} b t^{2}$

$v_{0} t + \frac{1}{2} b t^{2}$

$v_{0} t + \frac{1}{6} b t^{3}$

$v_{0} t + \frac{1}{3} b t^{3}$

Q. The acceleration of a particle is increasing linearly with time t as bt. The particle starts from origin with an initial velocity v0​. The distance travelled by the particle in time t will be

v0​t+31​bt2

v0​t+21​bt2

v0​t+61​bt3

v0​t+31​bt3

Acceleration ∝bt. i.e., dt2d2x​=a∝bt Integrating, dtdx​=2bt2​+C Initially, t=0,dx/dt=v0​ Therefore, dtdx​=2bt2​+v0​ Integrating again, x=6bt3​+v0​t+C When t=0,x=0 ⇒C=0. i.e., distance travelled by the particle in time t =v0​t+6bt3​.

Q. The acceleration of a particle is increasing linearly with time $t$ as $b t$ . The particle starts from origin with an initial velocity $v_{0}$ . The distance travelled by the particle in time t will be

$v_{0} t + \frac{1}{3} b t^{2}$

$v_{0} t + \frac{1}{2} b t^{2}$

$v_{0} t + \frac{1}{6} b t^{3}$

$v_{0} t + \frac{1}{3} b t^{3}$