Simple Kinematics in One Dimension – AIEEE 2008 & other questions

The following question which appeared in AIEEE 2008 question paper is an interesting one. You will find that you can work it out without the knowledge of any standard formula. In fact sheer imagination is sufficient to answer this question and that’s why it is interesting. Here is the question:

A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x₁(t) after time 't' and that of second body by x₂(t) after the same time interval. Which of the following graphs correctly describes (x₁ – x₂) as a function of time ‘t’?

(x₁ – x₂) must be negative in the initial stage since the first body starts from rest where as the second body has a constant speed through out its motion. The first body will be behind the second one and the separation between them will increase for some time. As the first body picks up speed, the separation between the bodies decreases and becomes zero at a certain instant. The separation (x₁ – x₂) will be negative up to this instant. Thereafter the first body overtakes the second one and the separation (x₁ – x₂) becomes positive and goes on increasing. The situation is indicated correctly by graph (1).

Alternatively, you can find the answer by applying the equation,

s = ut + (¹/₂)at² for uniformly accelerated one dimensional motion.

Thus x₁ = 0 + (¹/₂)at² and x₂ = vt where a is the constant acceleration of the first body and v is the constant speed of the second body.

Therefore, (x₁ – x₂) = (¹/₂)at² – vt

(x₁ – x₂) = 0 when t = 0 and when t = 2v/a.

Further, d(x₁ – x₂)/dt = at – v so that at t = v/a the quantity (x₁ – x₂) will be a maximum or minimum. It is in fact a minimum point on the curve obtained by plotting (x₁ – x₂) against t, as shown by the positive value of the second differential coefficient, d²(x₁ – x₂)/dt².

So the correct option is indeed curve (1).

Here is another question:

A particle starts from rest with a constant acceleration and moves along the positive x-direction. If its displacement is x₁ in the first two seconds and x₂ in the next two seconds, then

(a) x₂ = 2x₁

(b) x₂ = 3x₁

(c) x₂ = 4x₁

(d) x₂ = 5x₁

(e) x₂ = 6x₁

We have x₁ = (1/2) a×2² = 2a

x₂ = (1/2) a×4² – (1/2) a×2² = 6a

Therefore, x₂ = 3x₁

Now, consider the following question which is quite simple. But be careful while answering.

A train having a constant speed of 54 km/hour takes 10 seconds to move past a lamp post. How much time is required for this train to cross a 300 m long bridge?

(a) 18 s

(b) 20 s

(c) 30 s

(d) 36 s

(e) 40 s

The speed of the train in ms^–1 is 54×5/18 = 15.

Since the train takes 10 s to move past the lamp post, the length of the train is 15×10 = 150 m.

The train has to travel a total distance of 450 m to cross the bridge. (This includes the length of the train also).

Therefore, time required = 450/15 = 30 s.

The following question appeared in AIPMT 2008 question paper:

A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms^–1 to 20 ms^–1 while passing through a distance 135 m in t second. The value of t is

(1) 1.8

(2) 12

(3) 9

(4) 10

We have v_t ² = v₀ ² + 2as with usual notations.

Therefore, 20² = 10² + 2a×135 from which a = 30/27

Substituting this value of acceleration a in the equation

v_t = v₀ + at

we have 20 = 10 + (30/27)t from which t = 9 second

You will find some useful posts on one dimensional motion at physicsplus

Kerala Engineering Entrance 2008 Questions on One Dimensional Motion

The following questions (on one dimensional kinematics) numbered 1, 2 and 3 appeared in KEAM (Engineering) 2008 question paper:

(1) A particle starts from rest at t = 0 and moves in a straight line with an acceleration as shown below. The velocity of the particle at t = 3 s is

(a) 2 ms^–1

(b) 4 ms^–1

(c) 6 ms^–1

(d) 8 ms^–1

(e) 1 ms^–1

According to the acceleration – time graph shown, the particle has an acceleration of 4 ms^–2 during the first two seconds. Therefore, its velocity (v₂) at the end of 2 seconds is given by

v₂ = v₀ + at = 0 + 4×2 = 8 ms^–1

From 2 second to 3 second the particle has a retardation of 4 ms^–2. Hence its velocity (v₃) at the end of 3 seconds is given by

v₃ = v₂ – at = 8 – 4×1 = 4 ms^–1 [Option (b)].

(2) Two cars A and B are moving with same speed of 45 km/hr along same direction. If a third car C coming from the opposite direction with a speed of 36 km/hr meets two cars in an interval of 5 minutes, the distance of separation of two cars A and B should be (in km)

(a) 6.75

(b) 7.25

(c) 5.55

(d) 8.35

(e) 4.75

The relative velocity of car A with respect to car B (and that of car B with respect to A) is zero since they have the same speed in the same direction. The relative velocity of car C with respect to A and B is 45 + 36 = 81 km/hr. Since the car C takes a time of 5 minutes to cover the distance between A and B, the separation between A and B is 81×(5/60) km = 6.75 km.

(3) An object is dropped from rest. Its v - t graph is

The correct option is (a) since the velocity ‘v’ is directly proportional to the time t in accordance with the equation v = v₀ + at where v₀ is the initial velocity which is zero and a is the acceleration which is the constant acceleration due to gravity. The graph should evidently pass through the origin.

Suppose you were asked to draw the v – t graph in the case of a ball projected vertically upwards with a velocity u. If you are asked to draw the graph from the instant the ball leaves your hand to the instant you catch it while returning, you can do it as shown, ignoring the air resistance and the time taken for the velocity to reduce to zero on hitting your hand.

Multiple Choice Questions on One Dimensional Motion

Here is a question in kinematics which is a popular one and therefore requiring your attention:
The two ends of a train running with a constant acceleration passes a certain point with velocities v₁ and v₂. The velocity with which the middle point of the train passes the point is
(a) (v₁+v₂)/2 (b) √(v₁ ² +v₂²) (c) (v₁²+v₂²)/2 (d) (v₁+v₂)/√2

(e) √[(v₁²+v₂²)/2]
If ‘s’ is the length of the train, the velocity of the train changes from v₁ to v₂ when it moves through the distance ‘s’. Therefore we have,
v₂²- v₁² = 2as from which a = (v₂² - v₁² )/2s
If’ ’v’ is the velocity with which the mid point of the train passes the reference point, we have v² = v₁² +2a(s/2). Substituting for the acceleration ‘a’ from the above equation, v = √[(v₁² +v₂ ² )/2].
The following question appeared in the Kerala Engineering Entrance Test paper of 2002:
A body dropped from a height ‘h’ with an initial velocity zero reaches the ground with a velocity 3km/hour. Another body of the same mass is dropped from the same height ‘h’ with an initial velocity 4km/hour. It will reach the ground with a velocity
(a) 3km/hour (b) 4km/hour (c) 5km/hour (d) 12km/hour (e) 8km/hour
We have, v² = u² + 2as with usual notations.
Therefore, 3² = 0 + 2gh for the first case and
v² = 4² + 2gh for the second case. These two equations yield the value v= 5km/hour. Note that we did not convert the velocities into m/s since the answer is required in km/hour. [Note that the answer is independent of the masses of the bodies].
Now consider the following simple question:
A particle starting from rest travels with uniform acceleration for t₁ seconds and then travels (continuously) with uniform retardation and comes to rest in another t₂ seconds. If the total distance traveled is ‘s’, the maximum velocity attained during the motion is
(a) s/(t₁ – t₂) (b) s/(t₁ + t₂) (c) 2s/(t₁ – t₂) (d) 2s/(t₁ + t₂) (e) s/t₁ + s/t₂
Since the acceleration and retardation are uniform, this can be easily solved using the concept of average velocity. If ‘v’ is the maximum velocity attained, the average velocity during the acceleration part as well as the deceleration part is v/2.
Therefore, s = (v/2)t₁+ (v/2)t₂ = (v/2)(t₁+t₂) from which v = 2s/(t₁+t₂).

Minimum time of travel of an electric car- M.C.Q. from one dimensional motion

An experimental electric car can produce a maximum acceleration of 2m/s². Its brakes can produce a maximum retardation of 8m/s². What is the minimum time required by this electric car to start from rest and come to a stop, covering a distance of 500m?

(a) 10s (b) 15s (c) 20s (d) 25s (e) 30s

This question may appear to be a bit difficult at the first sight, but it is a simple one which can be solved using the basic equations of motion. The maximum velocity ‘v’ attained during the motion is given by

v² = 0+2a₁s₁ during the accelerated motion and

0 = v²-2a₂s₂ during the decelerated motion. From these equations, s₁ = v²/2a₁ and s₂ = v²/2a₂

Here s₁+ s_{2 =} 500m, a₁= 2m/s² and a₂ = 8m/s².

Therefore, 500 = v²/2a₁ + v²/2a_2. On substituting for a₁ and a_2, v = 40m/s.

But we have, v=0+a₁t₁ and 0=v-a₂t₂ for the accelerated and decelerated parts of the motion respectively so that t₁=v/a₁=40/2 = 20 s and t₂=v/a₂ = 40/8 = 5 s.

The total minimum time therefore is 25 s, given in option (d).