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.

Kerala Engineering Entrance (KEAM) 2011 Questions on Kinematics in One Dimension

“Common sense is the collection of prejudices acquired by age eighteen.”

– Albert Einstein

Today we will discuss three questions (MCQ) on one dimensional kinematics. These questions were included in the KEAM (Engineering) question paper. They are of the type often seen in similar entrance tests. Here are the questions with their solution:

(1) A bus begins to move with an acceleration of 1 ms^–2. A man who is 48 m behind the bus starts running at 10 ms^–1 to catch the bus. The man will be able to catch the bus after

(a) 6 s

(b) 5 s

(c) 3 s

(d) 7 s

(e) 8 s

If the man can catch the bus after t seconds we have

10 t = 48 + ½ at² where a is the acceleration of the bus (a = 1 ms^–2).

[10 t is the distance covered by the man in time t and ½ at² is the distance covered by the bus in the same time. Since the man is 48 m behind the bus, he has to cover an additional distance of 48 m].

Since a = 1 ms^–2 the above equation reduces to

t² – 20 t + 96 = 0

Therefore t = [–(–20) ±√{20² – (4×1×96}] /(2×1)

Or, t = [20 ±√16] /2 = 12 s or 8 s

The smaller time 8 s is the answer.

(2) A car moves a distance of 200m. It covers first half of the distance at speed 60 km h^–1 and the second half at speed v. If the average speed is 40 km h^–1, the value of v is

(a) 30 km h^–1

(b) 13 km h^–1

(c) 60 km h^–1

(d) 40 km h^–1

(e) 20 km h^–1

The total distance moved is 200 m = 0.2 km

The total time (in hours) taken by the car is (0.1/60) + (0.1/v)

Since the average velocity is 40 km h^–1, we have

40×[(0.1/60) + (0.1/v)] = 0.2

Or, 4/v = 0.2 – (4/60) = 8/60 from which v = 240/8 = 30 km h^–1

[If v₁ and v₂ are the velocities while traversing the two halves of the path length s, the average velocity v_av is given by

v_av = s/[(s/2v₁) + (s/2v₂)]

Or v_av = 2v₁v₂/( v₁+v₂)

Since this final equation contains velocities only you can make substitutions without confusion (no need of conversion of metre into kilometre when distances are to be handled).

Therefore, 40 = (2×60×v)/(60+v) from which v = 30 km h^–1]

(3) A particle is moving with constant acceleration from A to B in a straight line AB. If u and v are the velocities at A and B respectively then its velocity at the mid point C will be

(a) [(u² + v²)/2u]²

(b) (u + v)/2

(c) (v – u)/2

(d) √[(u² + v²)/2]

(e) √(v² – u²)/2

If ‘s’ is the length of the path AB, the velocity of the particle changes from u to v when it moves through the distance ‘s’. Therefore we have,

v² – u² = 2as from which the acceleration, a = (v² – u²)/2s

If’ ’v_c ’ is the velocity of the particle at the mid point C of the path AB, we have v_c² = u² + 2a(s/2). Substituting for the acceleration ‘a’ from the above equation, v_c = √[(u² + v²)/2].

Kinematics- KEAM (Engineering) 2009 Questions on One dimensional Motion

Today we will discuss some multiple choice questions on one dimensional motion which appeared in Kerala Engineering Entrance 2009 question paper:

(1) A ball is thrown up vertically with speed u and at the same instant another ball B is released from a height h. At time t, the speed of A relative to B is

(a) u

(b) 2u

(c) u – gt

(d) √(u² – gt)

(e) gt

If you know the concept of relative velocity correctly, this question will be quite simple for you.

The relative velocity of A with respect to B is given by

V_AB = V_A – V_B where V_A and V_B are respectively the velocities of A and B (with respect to the common frame of reference chosen).

The velocity of A and B at time t are respectively u – gt and – gt, taking upward quantities positive and downward quantities negative.

The speed of A relative to B is u – gt – (– gt) = u [Option (a)].

(2) A body is falling freely under gravity. The distances covered by the body in the first, second and third minutes of its motion are in the ratio

(a) 1 : 4 : 9

(b) 1 : 2 : 3

(c) 1 : 3 : 5

(d) 1 : 5 : 6

(e) 1 : 5 : 13

Since you require the ratio of distances, it is enough to consider the times in seconds (instead of minutes). Strictly, it must be mentioned that the body starts from rest. In the case of such a freely falling body, the distance covered is directly proportional to t² since s = ut + ½ gt² with usual notations where u = 0

The distances covered by the body in one second, two seconds and three seconds are in the ratio 1 : 4 : 9.

Therefore, the distances covered by the body in the first, second and third seconds are in the ratio 1 : (4 – 1) : (9 – 4) which is 1 : 3 : 5

(3) A bullet fired into a fixed wooden block loses half of its velocity after penetrating 40 cm. It comes to rest after penetrating a further distance of

(a) ²²/₃ cm

(b) ⁴⁰/₃ cm

(c) ²⁰/₃ cm

(d) ²²/₅ cm

(e) ²⁶/₅ cm

We have v² = u² + 2as where u is the initial velocity, v is the final velocity after suffering a displacement s and a is the acceleration.

Therefore we have

u²/4 = u² + 2a × 0.4

Or, – 3u²/4 = 2a × 0.4 …………..(i)

If the bullet comes to rest after penetrating a further distance of s₁ we have

0 = u²/4 + 2a × s₁

Or, – u²/4 = 2a × s₁ …………….(ii)

Dividing Eq(i) by Eq(ii) we get

3 = 0.4/s₁ from which s₁ = 0.4/3 m = ⁴⁰/₃ cm.

[The above question can be worked out using the work energy principle as well].

You will find similar useful multiple choice questions on kinematics (with solution) here.

AP Physics & Degree Entrance Kinematics- Questions on One Dimensional Motion

The following questions numbered 1, 2 and 3 are based on the velocity-time graph of a particle in one dimensional motion shown in the adjoining figure.

(1) The displacement of the particle during the first second of its motion is nearly

(a) 0.25 m

(b) 0.5 m

(c) 1 m

(d) 1.5 m

(e) 2 m

The area under the velocity-time curve gives the displacement. The portion of the graph for the interval from zero to 1 second is straight and the area under the curve is triangular and is equal to (½)×1×4 = 2 m.

(2) The average accelerations during the 3^rd second and 7^th second respectively are nearly

(a) 0.5 ms^–2 and 2 ms^–2

(b) 1 ms^–2 and –2 ms^–2

(c) 1 ms^–2 and 2 ms^–2

(d) 1 ms^–2 and –2 ms^–2

(e) zero and –2 ms^–2

The 3^rd second is the interval from 2 seconds to 3 seconds and during this time the velocity increases from 7 ms^–1 to 8 ms^–1. The acceleration is therefore (8 – 7)/1 = 1 ms^–2.

The 7^th second is the interval from 6 seconds to 7 seconds and during this time the velocity decreases from 8 ms^–1 to 6 ms^–1.

The acceleration is therefore (6 – 8)/1 = – 2 ms^–2. [Option (b)]

(3) Which one among the following acceleration–time graphs most closely represents the motion of the particle?

The acceleration is positive and uniform initially. Afterwards the acceleration becomes zero (since the velocity remains constant) for some time and then becomes negative (since the velocity goes on decreasing) and uniform. The curve shown in (b) therefore represents the motion of the particle.

Let us leave the velocity time graph here and consider a couple of different questions:

(4) Two boys running at uniform speeds v₁ and v₂ respectively along a straight line path in opposite directions get 9 m closer each second. While running along the same direction with their speeds reduced by 50%, they get 0.5m closer each second. The speeds v₁ and v₂ are respectively

(a) 6 ms^–1 and 3 ms^–1

(b) 5 ms^–1 and 4 ms^–1

(c) 5 ms^–1 and 4.5 ms^–1

(d) 5 ms^–1 and 3.5 ms^–1

(e) 5.5 ms^–1 and 3.5 ms^–1

This is a simple question involving relative velocity. Wile running along opposite directions, we have

v₁+ v₂ = 9

While running along the same direction, we have

v₁/2 – v₂/2 = 0.5, from which v₁– v₂ = 1

Solving the above equations, we obtain v₁ = 5 ms^–1 and v₂ = 4 ms^–1

(5) The rear end of a train running on a straight track with uniform acceleration has velocities 6 ms^–1 and 10 ms^–1 respectively when passing points A and B in its path. The velocity of the rear end midway between these points is approximately

(a) 7 ms^–1

(b) 7.5 ms^–1

(c) 8ms^–1

(d) 8.2 ms^–1

(e) 8.4 ms^–1

We have v² = u² + 2as where u and v are the initial ans final velocities respectively, a is the acceleration and s is the displacement.

If the distance between A and B is s, we have

10² = 6² + 2as from which 2as = 64

If v₁ is the velocity midway between A and B we have

v₁² = 6² + 2a(s/2) = 6² + 32 = 68

Therefore, v₁ = 8.2 ms^–1 nearly.

You will find more questions (with solution) on one dimensional motion and other sections at AP Physics Resources

Two Kerala Engineering Entrance 2007 Questions on Linear Motion

The following questions appeared in Kerala Engineering Entrance 2007 question paper:

(1) Two balls are dropped to the ground from different heights. One ball is dropped two seconds after the other but they both strike the ground at the same time. If the first ball takes 5 s to reach the ground, then the difference in initial heights is (g = 10 ms^–2)

(a) 20 m

(b) 80 m

(c) 170 m

(d) 40 m

(e) 160 m

The initial height of the first ball is the distance travelled by it in 5 seconds and is given by

x₁ = 0×5 + (½)×10×5², using the relation, x = v₀t + (½)at²

Therefore, x₁ = 125 m.

The time of travel of the second ball is 3 seconds and hence its initial height is given by

x₂ = 0×3 + (½)×10×3² = 45 m.

The difference in initial heights is x₁– x₂ =125 – 45 = 80 m.

(2) A ball is thrown vertically upwards with a velocity of 25 ms^–1 from the top of a tower of height 30 m. How long will it travel before it hits the ground?

(a) 6 s

(b) 5 s

(c) 4 s

(d) 12 s

(e) 10 s

Let us take all downward vectors positive. The displacement is 30 m and is positive. The acceleration due to gravity, g (equal to 10 ms^–2) also is positive. But, the velocity of projection is negative (being upwards).

From the equation, x = v₀t + (½)at², we have

30 = – 25 t + (½)×10×t²

Or, t² – 5 t – 6 = 0, from which t = [5±√(25+24)]/ 2 = + 6 s or –1 s.

Since negative time is impossible, the answer is 6 s.

[You could have taken upward quantities as positive if you wanted to, but you would get the same answer].