Questions from Kinematics (Including Karnataka CET 2008 Question)

“God used beautiful mathematics in creating the world.”

– P.A.M. Dirac

Today we shall discuss a few questions (MCQ) in the section, ‘kinematics in one dimension’. The questions I give you are meant for testing your knowledge, comprehension and the ability for applying what you have learned in this section.

(1) The velocity-time graphs of a car and a motor bike traveling along a straight road are shown in the adjoining figure. At time t = 0 they have the same position co-ordinate.  Pick out the correct statement from the following:

(a) At time t = 0 the car and the motor bike are at rest.

(b) At time t = 0 the car is at rest but the motor bike is moving.

(c) The distances traveled by the car and the motor bike in time t₁ are equal.

(d) The distance traveled by the bike in time t₁ is twice the distance traveled by the car in the same time.

(e) The distance traveled by the bike in time t₁ is half the distance traveled by the car in the same time.

You can easily conclude that options (a) and (b) are wrong.

To check the remaining options we use the equation of linear motion, s = ut + ½ at² where s is the displacement in time t, u is the initial velocity and a is the uniform acceleration.

The distance traveled by the car in time t₁ is v_ct₁ where v_c (let us say) is the constant velocity of the car.

[The velocity of the car is constant since the velocity-time graph of the car is parallel to the time axis].

The motion of the motor bike is uniformly accelerated and the acceleration a is given by

            a =  v_c/t₁

[Note that the velocity of the motor bike changes from 0 to v_c in time t₁.

The distance s traveled by the motor bike in time t₁ is given by

            s = ut + ½ at² = 0 + (½)×(v_c/t₁) t₁²

Or, s = v_ct₁/2

Therefore the distance traveled by the bike in time t₁ is half the distance traveled by the car in the same time [Option (e)].

(2) A student standing at the edge of a cliff throws a stone of mass m vertically upwards with speed v. It strikes the ground at the foot of the cliff with speed v₁. The student then throws another stone of mass m/4 vertically downwards with speed v. It strikes the ground at the foot of the cliff with speed v₂. If air resistance is negligible, v₁ and v₂ are related as

(a) v₁ = v₂

(b) v₁ = v₂/2

(c) v₁ = 2v₂

(d) v₁ = v₂/4

(e) v₁ = 4v₂

While returning, the sphere of mass m has downward speed v when it passes the edge of the cliff. For calculating the speed with which it strikes the ground at the foot of the cliff, the situation is similar to that of the stone of mass m/4 thrown downwards. Obviously both stones will strike the ground with the same speed [Option (a)].  

(3) A body is projected vertically upwards. The times corresponding to height h while ascending and descending are t₁ and t₂ are respectively. Then the velocity of projection is (g is acceleration due to gravity)

(1) g√(t₁t₂)

(2) gt₁t₂/(t₁+t₂)

(3) g√(t₁t₂)/2

(4) g(t₁+t₂)/2

The above question appeared in Karnataka CET 2008 question paper.

We have the following two equations to give h:

            h = ut₁ –  (½) g t₁²…………….(i)

            h = ut₂ –  (½) g t₂²…………….(ii)

(We have taken the displacement h and the initial velocity u (both upwards) as positive and that’s why the acceleration due to gravity g is negative).

(ii) – (i) gives u(t₂ – t₁) = (½) g(t₂² – t₁²)

Therefore u = (½) g(t₂² – t₁²)/(t₂ – t₁) = g(t₁+t₂)/2

You can find a few more multiple choice practice questions (with solution) in this section here.