J J Science Academy
1) A particle moves in a straight line and its position x at time t is given by $x=2+t$. Its velocity is
Hint:
$$x=2+t$$$$v=\frac{dx}{dt}=1=\text{constant}$$
2) The displacement ' x ' (in meter) of a particle of mass ' m ' (in kg ) moving in one dimension under the action of a force is released to time ' t ' (in sec) by $t=\sqrt{x}+3$. The displacement of the particle, when its velocity is zero will be
Hint:
$$t=\sqrt{x}+3$$$$\sqrt{x}=t-3$$$$x=(t-3)^2$$Differentiating equation (i) we get$$v=\frac{dx}{dt}=2(t-3)$$For the velocity of the particle will be zero, $t-3=0 \Rightarrow t=3 \text{ s}$The displacement of the particle, when its velocity is zero, is $x=(3-3)^2 \Rightarrow x=0$
3) The distance travelled by the particle along X -axis from 0 , is given by, $x=70+64 t-2 t^4$. It comes to rest at time $t=$
Hint:
$$x=70+64 t-2 t^4$$$$v=\frac{dx}{dt}=\frac{d}{dt}(70+64 t-2 t^4)=0+64-8 t^3$$Final velocity $=0$$64-8 t^3=0$$$$t^3=\frac{64}{8}=8$$$$t=2 \text{ s}$$
4) The motion of a particle is described by the equation, $x=a+bt^2$ where $a=15 \text{ cm/s}^2$ and $b=3 \text{ cm/s}^2$. Its instantaneous velocity at time 3 sec will be
Hint:
$$x=a+bt^2$$$$v=\frac{dx}{dt}=2bt$$At $t=3 \text{ s}$, $v=2 \times 3 \times 3=18 \text{ cm/s}$
5) A particle is moving in a straight line such that is displacement is given by, $s=\frac{t^4}{4}-2 t^3+6 t^2+15$, where $s$ is in metre and $t$ in second. The average acceleration during time interval $t=0$ to $t=4 \text{ s}$ is
Hint:
$$s=\frac{t^4}{4}-2 t^3+6 t^2+15$$$$v=\frac{ds}{dt}=t^3-6 t^2+12 t$$$$t_1=0, v_1=0$$$$t_2=4 \text{ s}, v_2=64-96+48=16 \text{ m/s}$$$$\overline{a}=\frac{v_2-v_1}{t_2-t_1}=\frac{16-0}{4-0}=4 \text{ m/s}^2$$
6) The displacement $x$ of a particle along a straight line at time $t$ is given by, $x=2+3 t+5 t^2$. The acceleration of the particle at $t=2 \text{ s}$ is
Hint:
$$x=2+3 t+5 t^2$$$$v=\frac{dx}{dt}=3+10 t$$$$a=\frac{dv}{dt}=10=\text{constant}$$
7) A particle is moving along x -axis such that velocity and displacement are related as $v=\alpha x^{1/2}$, the acceleration of particle will be
Hint:
$$v=\frac{dx}{dt}=\alpha x^{1/2}$$$$\frac{dv}{dt}=\alpha \frac{1}{2} x^{-1/2} \frac{dx}{dt}=\frac{\alpha}{2} x^{-1/2}(\alpha x^{1/2})$$$$a=\frac{\alpha^2}{2}$$
8) If $v$ is the velocity of a body moving along $x$-axis is $v=-2 t^2+t-3$, then acceleration of body at $t=4 \text{ s}$, is
Hint:
$$v=-2 t^2+t-3$$$$a=\frac{dv}{dt}=-4 t+1$$At $t=4 \text{ s}, a=-15$As velocity of a body moving along x -axis, acceleration is also along negative x -axis.
9) The displacement ' x ' of a particle at any instant is related to its velocity as, $v=\sqrt{2 x+9}$. Its acceleration is
Hint:
$$v=\sqrt{2 x+9}=(2 x+9)^{1/2}$$$$a=\frac{dv}{dt}=\frac{1}{2}(2 x+9)^{-1/2} \frac{d}{dt}(2 x)$$$$=\frac{1}{2 \sqrt{2 x+9}} 2 \frac{dx}{dt}$$$$=\frac{v}{\sqrt{2 x+9}}$$$$=\frac{\sqrt{2 x+9}}{\sqrt{2 x+9}}=1 \text{ unit}$$
10) The equation of motion of a body is given by, $x=t^3-3 t^2+12$. Its acceleration, at the beginning, is
Hint:
$$x=t^3-3 t^2+12$$$$v=\frac{dx}{dt}=3 t^2-6 t$$$$a=\frac{dv}{dt}=6 t-6$$At $t=0$, $a=-6 \text{ m/s}^2$
11) The position $x$ of a body as a function of time $t$ is given by the equation, $x=2 t^3-12 t^2+24 t+6$. The direction of motion of the body changes at $t=$
12) A particle moves a distance x in time t according to equation $x=(t+5)^{-1}$. The acceleration of particle is proportional to
Hint:
$$x=\frac{1}{t+5}$$Differentiating eq. (i) w.r.t. t, we get$$v=\frac{dx}{dt}=\frac{-1}{(t+5)^2}$$Differentiating eq. (ii) w.r.t. t, we get$$a=\frac{dv}{dt}=\frac{2}{(t+5)^3}$$Comparing eqs. (ii) and (iii), we get $a \propto v^{3/2}$
13) A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $v(x)=\beta x^{-2n}$, where $\beta$ and $n$ are constants and $x$ is the position of the particle. The acceleration of the particle as a function of $x$ is given by
Hint:
$$v(x)=\beta x^{-2n}$$$$a=v \frac{dv}{dx}=\left[\beta x^{-2n}\right] \frac{d\left(\beta x^{-2n}\right)}{dx}$$$$=\beta x^{-2n}\left[\beta(-2n) x^{-2n-1}\right]$$$$=-2 n \beta^2 x^{-2n-1-2n}=-2 n \beta^2 x^{-4n-1}$$
14) A particle moves in a straight line, its position (in m ) as function of time is given by $x=(at)^2+b$. What is average velocity in time interval $t=3 \text{ sec}$ to $t=5 \text{ sec}$? (where $a$ and $b$ are constants and $a=1 \text{ m/s}^2, b=1 \text{ m}$)
Hint:
$$t_1=3 \text{ s}$$$$x_1=[1 \times(3)^2]+1=10$$$$t_2=35 \text{ s}$$$$x_2=[1 \times(5)^2]+1=26$$$$v_{av}=\frac{x_2-x_1}{t_2-t_1}=\frac{26-10}{5-3}=\frac{16}{2}=8 \text{ m/s}$$
15) The position of a particle moving on x -axis is given by, $x=At^3+Bt^2+Ct+D$. The numerical value of $A, B, C, D$ are $1,4,-2$ and $5$ respectively and S.I. units are used. What is the velocity of the particle at $t=4 \text{ s}$?
Hint:
$$x=At^3+Bt^2+Ct+D$$$$v=\frac{dx}{dt}=3 At^2+2 Bt+C$$At $t=4 \text{ s}$$$$v=(3 \times 1 \times 16)+(2 \times 4 \times 4)+(-2)=48+32-2=78 \text{ m/s}$$
16) A particle is moving in a straight line under retardation $a=\lambda v$, where $\lambda$ is constant. If $v_0$ is the initial velocity, after how much time the velocity of particle will be $v_0/4$?
Hint:
$$a=\frac{dv}{dt}=-\lambda v$$$$\int_{v_0}^{v} \frac{dv}{v}=-\lambda \int_{0}^{t} dt$$$$v=v_0 e^{-\lambda t}$$$$\frac{v_0}{4}=v_0 e^{-\lambda t}$$$$e^{\lambda t}=4$$$$t=\frac{\ln 4}{\lambda}=\frac{2 \ln 2}{\lambda}$$
17) The acceleration a (in $ms^{-2}$ ) of a body, starting from rest varies with time t (in s) following the equation, $a=3 t+4$. The velocity of the body at time $t=2 \text{ s}$ will be
Hint:
$$a=\frac{dv}{dt}=3 t+4$$$$\int_{0}^{v} dv=\int_{0}^{2}(3 t+4) dt=\left[\frac{3 t^2}{2}+4 t\right]_{0}^{2}=6+8=14 \text{ m/s}$$
18) If the velocity of a particle is $v=At+Bt^2$, where A and B are constants, then the distance travelled by it between $1 \text{ s}$ and $2 \text{ s}$ is
Hint:
$$v=\frac{dx}{dt}=At+Bt^2$$$$\int_{0}^{x} dx=\int_{1}^{2}(At+Bt^2) dt=\left[\frac{At^2}{2}+\frac{Bt^3}{3}\right]_{1}^{2}$$$$x=\frac{A}{2}[(2)^2-(1)^2]+\frac{B}{3}[(2)^3-(1)^3]=\frac{3}{2} A+\frac{7}{3} B$$
19) A particle starts from rest. Its acceleration (a) versus time (t) graph is as shown in the figure. The maximum speed of the particle will be
Hint:
The area under the acceleration-time graph gives change in velocity. Since, particle starts with $u=0$, therefore$$\text{Change in velocity}=\text{area under a-t graph}$$$$v_{max}-0=\frac{1}{2} \times 10 \times 11$$$$v_{max}-0=55 \text{ m/s}$$