# PHYSICS – KINEMATICS

** Physics for Engineering/Medical**

** Kinematics**

**Question – 1. **

** The velocity time graph of a particle is represented by y = mx + c, the particle is moving with : **

**(A) ****Constant velocity (C) Constant acceleration **

**(B) ****Variable velocity (D) Constant momentum **

**Solution : –**

We have, the velocity –time graph has positive slope. Hence, the acceleration is constant.

Ans is (C) **[Ans.] **

**Question – 2. **

** An aeroplane flying at a constant speed releases a bomb. When the bomb moves away from the aeroplane, it will : **

**(A) ****Always be vertically below the aeroplane, if the aeroplane is flying horizontally. **

**(B) ****Always be vertically below the aeroplane.**

**(C) **** Always be vertically below the aeroplane, if the aeroplane is flying at an angle of 35˚ to the horizon.**

**(D) **** Gradually falls behind the aeroplane. **

**Solution : –**

We have, the horizontal component of the velocity of the bomb and the aeroplna are the same. Hence, horizontal displacements would remain the same at all time. (A) **[Ans.]**

**Question – 3. **

**A car travels 150 km east and then 150 km south. Finally, it comes back to the starting point by the shortest route. The speed is 60 kmh ^{–1} throughout the journey. The average velocity is : **

**(A) ****120 kmh ^{–1} (B) 90 kmh^{–1} (C) 60 kmh^{–1} (D) 0 kmh^{–1} **

**Solution : –**

We have, the net displacement zero; hence average velocity zero (D) **[Ans.]**

**Question – 4. **

**The x and y coordinates of a particle at any time t are given by x = 2t + 4t ^{2} and y = 5t. The aceeleration of the particle at t = 5 is given by : **

**(A) ****Zero (B) 8 ms ^{–2} (C) 16 ms^{–2} (D) 32 ms^{–2} **

**Solution : –**

We have, v_{x} = d/dt(2t + 4t^{2}) = 2 + 8t

a_{x} = d/dt(2 + 8t) = 8

v_{y} = d/dt(5t) = 0

a_{y} = d/dt(v_{y}) = 0

The acceleration is along x-axis and is 8 ms^{–2} (B) **[Ans.] **

**Question – 5. **

**A projectile is thrown at an angle of β with the vertical. It reaches a maximum height H. The time taken to reach the highest point of its path is : **

**(A) ****√(H/g) (B) √(H/2g) (C) √(2H)/g (D)√(2H/cosβ)**

**Solution : –**

We have, H = (v2cos^{2}β)/(2g)

Or, vcosβ = √(2gH)

And t = (vcosβ)/g

= √(2gH)/g

= √(2gH)/g (C) **[Ans.]**

**Question – 6. **

**The equation of a projectile is as **

**y = x√3 – (gx ^{2})/2. **

**The angle of projection is : **

**(A) **** 15˚ (B) 30˚ (C) 45˚ (D) 60˚**

**Solution : –**

We have, by comparing

y = x√3 – (gx^{2})/2 and y = x tan θ – (gx^{2})/(2v^{2}cos^{2}θ),

tan θ = √3

Or, θ = 60˚ (D) **[Ans.]**

**Question – 7. **

**A particle makes n revolution in one second in a circular orbit of radius r. Its centripetal acceleration would be : **

**(A) ****4πn ^{2}r^{2} (B) 4π^{2}n^{2}r (C) 4π^{2}nr (D) 4πnr^{2} **

**Solution : –**

We have, ac = rω^{2}

= r(2πn)^{2}

= 4π^{2}n^{2}r (B) **[Ans.] **

**Question – 8. **

**A motion in a straight line is supplied with a constant power. The distance travelled in time t would be proportional to : **

**(A) ****t ^{3/2} (B) t ^{–3/2} (C) t ^{–1/2} (D) t^{1/2} **

**Solution : –**

We have, Power (P) = (1/2)[(mv^{2})/t]

Or, v^{2} = (2Pt)/m

Or, v = √[(2Pt)/m]

Or, dx/dt = [√{(2P)/m}]t^{1/2}

Integrating, we get x = [√ {(2P)/m}][(t^{3/2})(3/2)]

Or, x α t^{3/2} (A) **[Ans.] **

**Question – 9. **

**Displacement x = t ^{4}, the ratio of acceleration and velocity would be : **

**(A) ****t/3 (B) t/4 (C) 3t (D) 3/t **

**Solution : – **

We have, x = t^{4}

Therefore, velocity = dx/dt = 4t^{3}

And acceleration = d^{2}x/dt^{2 } = 12t^{2}

Therefore, Acceleraton/Velocity = (12t^{2})/(4t^{3})

= 3/t (D) **[Ans.] **

**Question – 10. **

**A projectile is projected and it is found that the horizontal range and the maximum height are equal. The angle of projection would be : **

**(A) ****θ = tan ^{–1} (4) (B) θ = tan^{–1} (3) (C) θ = tan^{–1} (2) (D) θ = tan^{–1} (1) **

**Solution : –**

We have, Horizontal range = (u^{2} sin 2θ)/g

Maximum height = (u^{2} sin^{2} θ)/(2g)

Both are equal

Therefore, (u^{2} sin^{2} θ)/(2g) = (u^{2} sin 2θ)/g

Or, sin^{2} θ/2 = 2 sin θ cos θ

Or, tan θ = 4 (A) **[Ans.] **

**Questiopn – 11. **

**A constant force directed towards a fixed point is applied to a body in motion. The magnitude of the force varies inversely as the square of the distance from the fixed point. The path of the motion would be : **

**(A) ****Parabola (B) Hyperbola (C) Circle (D) Ellipse **

**Solution : –**

We have, the force is directed towards a fixed point and changes the direction of motion without changing its speed. Hence, the path of the motion is a circle. (C) **[Ans.]**

**Question – 12. **

**A particle performing uniform circular motion with constant speed v in a circle of radius r. The tangential acceleration would be : **

**(A) ****Zero (B) v r (C) v ^{2} r (D) v^{2}/r **

**Solution : –**

We have, a particle performing uniform circular motion has radial acceleration only and no tangential acceleration. (A) **[Ans.]**

**Question – 13. **

**The ratio of time of ascent to the time of descent of a projectile projected at an angle of 50˚ with the horizontal is : **

**(A) ****1 : 2 (B) 2 : 3 (C) 3 : 2 (D) 1 : 1 **

**Solution: –**

We have, in a projectile,

Time of ascent = Time of descent

= (u sin θ)/g

Therefore, Time of ascent/Time of descent = [(u sin θ)/g]/[(u sin θ)/g]

= 1 : 1 (D) **[Ans.]**

**Question – 14. **

**A ball of mass m performs uniform circular motion. The radius of circular path is R. The linear momentum is represented by P. The radial force acting on the particle would be : **

**(A) ****P ^{2}mR (B) P^{2}m/R (C) P^{2}/mR (D) P^{2}R/m **

**Solution : –**

We have, Momentum, P = mv — — — (1)

Radial force, F = mv^{2}/R — — — (2)

Putting, v = P/m in (2), we get

F = (m/R)(P/m)^{2}

= P^{2}/mR (C) **[Ans.] **

**Question – 15. **

**A particle excutes circular motion under a central attractive force inversely proportional to distance R. The speed of the particle would be : **

**(A) ****Dependent on R (C) Dependent on R ^{2} **

**(B) ****Independent on R (D) Independent on 1/R **

**Solution : –**

We have, Centripetal force, F = mv^{2}/R

Therefore, F α 1/R

Therefore, v does not depend on R. (B) **[Ans.] **

**Question – 16. **

**A particle of mass m excutes a circular motion in a circular path of radius R. The centripetal acceleration varies as square of time given by βRt ^{2}. The force acting on the particle provides a power given by : **

**(A) ****mβR ^{2}t (B) mβR^{2}t^{2} (C) mβRt^{2} (D) mβrt**

**Solution : –**

We have, tangential acceleration exists, hence power is required.

Acceleration = v^{2}/R = βRt^{2}

Or, v = Rt√β

Tangential acceleration = a_{T} = dv/dt

Or, a_{T} = d(Rt√β)/dt

= R√β

Force = ma_{T} = mR√β

Power = force×velocity

= (mR√β)×(Rt√β)

= mβR^{2}t (A) **[Ans.]**

**Question – 17. **

**A projectile is projected at an angle θ with horizontal. When the particle makes an angle β with the horizontal, its speed v is given by : **

**(A) ****v = u cos θ cos β (B) u cos θ sin β (C) u sin θ cos β (D) u cos θ sec β**

**Solution : –**

We have, as horizontal component of velocity remains constant.

Therefore, u cos θ = v cos β

Or, v = u cos θ sec β (D) **[Ans.]**

**Question – 18. **

**A projectile is projected from a point and the point is choosen as origin. In t seconds, vertical distance y = 8t – 15 t ^{3} and horizontal distance x = 6t – 13t^{3}. The velocity of projection of projectile is given by : **

**(A) ****8 m/s (B) 10 m/s (C) 12 m/s (D) 14 m/s. **

**Solution : **–

We have, y = 8t – 15 t^{3}

dy/dt = 8 – 45 t^{2}

At, t = 0, v_{y} = 8

x = 6t – 13 t^{3}

dx/dt = 6 – 39 t^{2}

At, t = 0, v_{x} = 6

(Velocity)^{2} = v_{y}^{2} + v_{x}^{2}

= 8^{2} = 6^{2}

= 100

Velocity = 10. (B) **[Ans.] **

**Question – 19. **

**A particle describes parabolic paths in time t _{1} and t_{2} and covers the same horizontal range R. Then t_{1}t_{2} is proportional to : **

**(A) ****R ^{3} (B) R^{2} (C) R (D) √R**

**Solution : –**

We have, Time of flight = (2u sin θ/g

Horizontal range = (u^{2} sin 2θ)/g

As, the range is the same, angle of projection should be θ and 90˚ – θ.

Therefore, t_{1} = (2u sin θ)/g

And t_{2} = {2u sin (90˚ – θ)}/g

= (2u cos θ)/g

Therefore, t_{1}t_{2} = [(2u sin θ)/g]×[(2u cos θ)/g]

= (4u^{2} sin θ cos θ)/g2

= 2[(u^{2} sin 2θ)/g]/g

= 2R/g

Hence, t_{1}t_{2} α R (C) **[Ans.]**

**Question – 20. **

**If a car covers 2/5 of the total distance with speed v1 and 3/5 distance with speed v2 then the average speed is : **

**(A) ****(v _{1} + v_{2})/2 (C) 5v_{1}v_{2}/(3v_{1} + 2v_{2}) (C)2v_{1}v_{2}/(v_{1} + v_{2}) (D) v_{1}v_{2} **

**Solution : –**

We have, Speed = Distance/Time

Time = Distance/Speed

Let total distance = S,

Total time = t_{1} + t_{2} = (2S/5)/v_{1}+(3S/5)/v_{2}

Therefore, Average velocity = S/{(2S/5)/v_{1}+(3S/5)/v_{2}}

= 5v_{1}v_{2}/(3v_{1} + 2v_{2}) (B) **[Ans.]**

**Question – 21. **

**The acceleration of a particle is increasing linearly with time t as at. The particle starts from the origin with an initial speed of u. The distance travelled by the particle in time t would be : **

**(A) ****ut + 1/6 at ^{3} (B) ut + 1/4 at^{3} (C) ut + 1/3 at^{3} (D) ut + 1/2 at^{2} **

**Solution : –**

We have, acceleration = dv/dt = at

Therefore, dv = atdt

Therefore, _{u}∫^{v}dv = _{0}∫^{t}atdt

Or, v – u = at^{2}/2

Or, v = u + at^{2}/2 = ds/dt

Or, ds = udt + (at^{2}/2)dt

Integrating, we get s = ut + 1/6 at^{3} (A) **[Ans.]**

**Question – 22. **

**A particle moves in a straight line so that its displacement x in meter at time t in second is given by x ^{2} = 1 + t^{2}. The acceleration of the particle in m/sec^{2} at time t is given by : **

**(A) ****t/x ^{2} (B) 1/x – t/x^{2} (C) 1/x – t^{2}/x^{3} (D) t^{2}/x^{2}**

**Solution : –**

We have, x^{2} = 1 + t^{2}

Or, x = (1 + x^{2})^{1/2}

Therefore, dx/dt = (1/2) × (1 + t^{2})^{–1/2} × 2t

= t(1 + t^{2})^{–1/2}

Acceleration = d^{2}x/dt^{2} = t × (–1/2) × (1 + t^{2})^{–3/2}× 2t + (1 + t^{2})^{–1/2}

= 1/x – t^{2}/x^{3} (C) **[Ans.] **

**Question – 23. **

**A projectile is projected with an initial speed of (xi + yj ) m/sec. If the range of the projectile is double the maximum height reached by it then : **

**(A) ****x = y (B) y = 2x (C) y = 3x (D) y = 4x. **

**Solution : –**

We have, u sin θ = y and u cos θ = x.

Therefore, H = (u^{2} sin^{2} θ)/2g = y^{2}/2g

And R = [^{u2}(2 sin θ cos θ)]/g = 2xy/g

It is given that R = 2H

Therefore, 2xy/g = 2y^{2}/2g

Or, y = 2x (B) **[Ans.] **

**Question – 24. **

**A stone tied to the end a string of 1 m long is whirled in a horizontal plane in a circle with a constant speed. The stone makes 30 revolution in 1 minute. The magnitude and direction of acceleration of the stone would be : **

**(A) ****π ^{2} ms^{–2} along the radius toward the centre**

**(B) ****π ^{2} ms^{–2} along the tangent to the circle **

**(C) ****π ^{2} ms^{–2} along the radius away from the centre **

**(D) ****π ^{2} ms^{–3} away from the centre. **

**Solution : –**

We have, a = rω^{2}; ω = 2πν

Again 30 revolution = 60 sec

Or, 1 revolution = 2 sec

Therefore, ν = 1/2 Hz

And a = rω^{2}

= 1×(4π^{2})/4

= π^{2} ms^{–2} (A) & (C) **[Ans.]**

**Question – 25. **

**The speed of a boat in still water is u and that in river is v. The time taken by the boat to reach a place downstream at a distance d and come back to the original place is : **

**(A) ****ud/(u ^{2} – v^{2}) (B) 2ud/(u^{2} – v^{2}) (C) 2vd/(u^{2} – v^{2}) (D) vd/(u^{2} – v^{2}) **

**Solution : –**

We have, the velocity downstream = u + v and velocity upstream = u – v. (u > v), otherwise boat will not return back upstream.

Therefore, Time for to and fro trip = d/(u + v) + d/(u – v)

= 2ud/(u^{2} – v^{2}) (B) **[Ans.]**

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