Paper Questions (Raw)
Q1. A particle moves in a straight line and its position x at time t is given by ($x=2+t$). Its velocity is
1) constant
2) ($\propto t$)
3) ($\propto t^2$)
4) ($\propto t^{-1}$)
Hint: $$x=2+t$$$$v=\frac{dx}{dt}=1=\text{constant}$$
Q2. 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
1) 2 m
2) 4 m
3) zero
4) 6 m
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 ($x=(3-3)^2 \Rightarrow x=0$)
Q3. 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 ($t=$)
1) 2 s
2) 1.85 s
3) 5 s
4) 2.87 s
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 ($=0$)$$64-8 t^3=0$$$$t^3=\frac{64}{8}=8$$$$t=2 \text{ s}$$
Q4. 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
1) ($36 \text{ cm/sec}$)
2) ($18 \text{ cm/sec}$)
3) ($16 \text{ cm/sec}$)
4) ($12 \text{ cm/sec}$)
Hint: $$x=a+bt^2$$$$v=\frac{dx}{dt}=2bt$$At ($t=3 \text{ s}$), ($v=2 \times 3 \times 3=18 \text{ cm/s}$)
Q5. 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
1) ($1 \text{ m/s}^2$)
2) ($2 \text{ m/s}^2$)
3) ($3 \text{ m/s}^2$)
4) ($4 \text{ m/s}^2$)
Hint: $$s=\frac{t^4}{4}-2 t^3+6 t^2+15$$$$v=\frac{ds}{dt}=t^3-6 t^2+12 t$$$$t_1=0, v_1=0$$$$t_2=4 \text{ s}, v_2=64-96+48=16 \text{ m/s}$$$$\overline{a}=\frac{v_2-v_1}{t_2-t_1}=\frac{16-0}{4-0}=4 \text{ m/s}^2$$
Q6. 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: $$x=2+3 t+5 t^2$$$$v=\frac{dx}{dt}=3+10 t$$$$a=\frac{dv}{dt}=10=\text{constant}$$
Q7. 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
1) ($\frac{\alpha}{2}$)
2) ($\frac{\alpha^2}{2}$)
3) ($\alpha^2 x$)
4) ($\alpha x^{3/2}$)
Hint: $$v=\frac{dx}{dt}=\alpha x^{1/2}$$$$\frac{dv}{dt}=\alpha \frac{1}{2} x^{-1/2} \frac{dx}{dt}=\frac{\alpha}{2} x^{-1/2}(\alpha x^{1/2})$$$$a=\frac{\alpha^2}{2}$$
Q8. 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
1) ($-15 \hat{i}$)
2) ($15 \hat{i}$)
3) ($-30 \hat{j}$)
4) ($-15 \hat{j}$)
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.
Q9. The displacement ' x ' of a particle at any instant is related to its velocity as, ($v=\sqrt{2 x+9}$). Its acceleration is
1) 1 unit
2) 2 unit
3) 0.5 unit
4) 4 unit
Hint: $$v=\sqrt{2 x+9}=(2 x+9)^{1/2}$$$$a=\frac{dv}{dt}=\frac{1}{2}(2 x+9)^{-1/2} \frac{d}{dt}(2 x)$$$$=\frac{1}{2 \sqrt{2 x+9}} 2 \frac{dx}{dt}$$$$=\frac{v}{\sqrt{2 x+9}}$$$$=\frac{\sqrt{2 x+9}}{\sqrt{2 x+9}}=1 \text{ unit}$$
Q10. The equation of motion of a body is given by, ($x=t^3-3 t^2+12$). Its acceleration, at the beginning, is
1) ($6 \text{ m/s}^2$)
2) 0
3) ($-6 \text{ m/s}^2$)
4) ($8 \text{ m/s}^2$)
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 ($t=0$), ($a=-6 \text{ m/s}^2$)
Q11. 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 ($t=$)
1) 1 s
2) 2 s
3) 3 s
4) 0.5 s
Q12. A particle moves a distance x in time t according to equation ($x=(t+5)^{-1}$). The acceleration of particle is proportional to
1) ($(\text{velocity})^{3/2}$)
2) ($(\text{distance})^2$)
3) ($(\text{distance})^{-2}$)
4) ($(\text{velocity})^{2/3}$)
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 ($a \propto v^{3/2}$)
Q13. 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
1) ($-2 n \beta^2 x^{-2n-1}$)
2) ($-2 n \beta^2 x^{-4n-1}$)
3) ($-2 n \beta^2 x^{-2n+1}$)
4) ($-2 n \beta^2 x^{-4n+1}$)
Hint: $$v(x)=\beta x^{-2n}$$$$a=v \frac{dv}{dx}=\left[\beta x^{-2n}\right] \frac{d\left(\beta x^{-2n}\right)}{dx}$$$$=\beta x^{-2n}\left[\beta(-2n) x^{-2n-1}\right]$$$$=-2 n \beta^2 x^{-2n-1-2n}=-2 n \beta^2 x^{-4n-1}$$
Q14. 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 ($a=1 \text{ m/s}^2, b=1 \text{ m}$))
1) ($8 \text{ m/s}$)
2) ($5 \text{ m/s}$)
3) ($10 \text{ m/s}$)
4) ($12 \text{ m/s}$)
Hint: $$t_1=3 \text{ s}$$$$x_1=[1 \times(3)^2]+1=10$$$$t_2=35 \text{ s}$$$$x_2=[1 \times(5)^2]+1=26$$$$v_{av}=\frac{x_2-x_1}{t_2-t_1}=\frac{26-10}{5-3}=\frac{16}{2}=8 \text{ m/s}$$
Q15. 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 ($t=4 \text{ s}$)?
1) ($78 \text{ m/s}$)
2) ($87 \text{ m/s}$)
3) ($68 \text{ m/s}$)
4) ($97 \text{ m/s}$)
Hint: $$x=At^3+Bt^2+Ct+D$$$$v=\frac{dx}{dt}=3 At^2+2 Bt+C$$At ($t=4 \text{ s}$)$$v=(3 \times 1 \times 16)+(2 \times 4 \times 4)+(-2)=48+32-2=78 \text{ m/s}$$
Q16. 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 ($v_0/4$)?
1) ($\frac{\ln 2}{\lambda}$)
2) ($\frac{2 \ln 2}{\lambda}$)
3) ($\frac{3 \ln 2}{\lambda}$)
4) ($\frac{\ln 2}{2 \lambda}$)
Hint: $$a=\frac{dv}{dt}=-\lambda v$$$$\int_{v_0}^{v} \frac{dv}{v}=-\lambda \int_{0}^{t} dt$$$$v=v_0 e^{-\lambda t}$$$$\frac{v_0}{4}=v_0 e^{-\lambda t}$$$$e^{\lambda t}=4$$$$t=\frac{\ln 4}{\lambda}=\frac{2 \ln 2}{\lambda}$$
Q17. 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
1) ($10 \text{ m/s}$)
2) ($18 \text{ m/s}$)
3) ($14 \text{ m/s}$)
4) ($26 \text{ m/s}$)
Hint: $$a=\frac{dv}{dt}=3 t+4$$$$\int_{0}^{v} dv=\int_{0}^{2}(3 t+4) dt=\left[\frac{3 t^2}{2}+4 t\right]_{0}^{2}=6+8=14 \text{ m/s}$$
Q18. 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
1) ($\frac{3}{2} A+4 B$)
2) ($3 A+7 B$)
3) ($\frac{3}{2} A+\frac{7}{3} B$)
4) ($\frac{A}{2}+\frac{B}{3}$)
Hint: $$v=\frac{dx}{dt}=At+Bt^2$$$$\int_{0}^{x} dx=\int_{1}^{2}(At+Bt^2) dt=\left[\frac{At^2}{2}+\frac{Bt^3}{3}\right]_{1}^{2}$$$$x=\frac{A}{2}[(2)^2-(1)^2]+\frac{B}{3}[(2)^3-(1)^3]=\frac{3}{2} A+\frac{7}{3} B$$
Q19. 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
1) ($110 \text{ m/s}$)
2) ($55 \text{ m/s}$)
3) ($550 \text{ m/s}$)
4) ($660 \text{ m/s}$)
Hint: The area under the acceleration-time graph gives change in velocity. Since, particle starts with ($u=0$), therefore$$\text{Change in velocity}=\text{area under a-t graph}$$$$v_{max}-0=\frac{1}{2} \times 10 \times 11$$$$v_{max}-0=55 \text{ m/s}$$