The acceleration of a particle is increasing | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) $\frac{{dv}}{{dt}} = bt \Rightarrow dv = bt\;dt \Rightarrow v = \frac{{b{t^2}}}{2} + {K_1}$

At $t = 0,\;v = {v_0} \Rightarrow {K_1} = {v_0}$

Vedclass · Accepted Answer

${v_0}t + \frac{1}{6}b{t^3}$

Vedclass · Answer

(c) $\frac{{dv}}{{dt}} = bt \Rightarrow dv = bt\;dt \Rightarrow v = \frac{{b{t^2}}}{2} + {K_1}$

At $t = 0,\;v = {v_0} \Rightarrow {K_1} = {v_0}$

We get $v = \frac{1}{2}b{t^2} + {v_0}$

Again $\frac{{dx}}{{dt}} = \frac{1}{2}b{t^2} + {v_0} \Rightarrow x = \frac{1}{2}\frac{{b{t^3}}}{3} + {v_0}t + {K_2}$

At $t = 0,\;x = 0 \Rightarrow {K_2} = 0$

$	herefore x = \frac{1}{6}b{t^3} + {v_0}t$

The acceleration of a particle is increasing linearly with time $t$ as $bt$. The particle starts from the origin with an initial velocity ${v_0}$. The distance travelled by the particle in time $t$ will be

Similar Questions

The $v - t$ graph of a moving object is given in figure. The maximum acceleration is...........$\mathrm{cm/sec}^{2}$

The velocity-displacement graph of a particle is shown in the figure.

The acceleration-displacement graph of the same particle is represented by :

For a moving body at any instant of time

The velocity-time and acceleration-time graphs of a particle are given as Its position-time graph may be gvien as