Motion in a Straight Line 2 - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 14 Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Motion in a Straight Line – 2”.

1. A particle moving along a straight line with uniform acceleration, passes successively through three points A, B, C. If t₁ = time taken to go from A to B; t₂ = time taken to go from B to C and AB = a, BC = b. If a = b and the velocities of the particle at A, B, C be p, q, r respectively, then, what is the relation between p², q² and r²?
a) A.P
b) G.P
c) H.P
d) They are in any arbitrary series
View Answer

Answer: a
Explanation: When a = b, then the equation of the motion of the particle from A to B is,
q² = p² + 2fa
Or q² – p² = 2fa ……….(1)
Again, the equation of motion of the particle from B to C is,
r² = q² + 2fa
Or r² – q² = 2fa ……….(2)
From (1) and (2) we get,
q² – p² = r² – q²
Thus, p², q² and r² are in A.P.

2. Two motor cars on the same line approach each other with velocities u₁ and u₂ respectively. When each is seen from the other, the distance between them is x. If f₁ and f₂ to be the maximum retardation of the two cars then a collision can be just avoided then at which condition collision can just be avoided?
a) (u₁²f₂ – u₂²f₁) = 2f₁f₂(x)
b) (u₁²f₂ + u₂²f₁) = 2f₁f₂(x)
c) (u₁²f₂ + u₂²f₁) = f₁f₂(x)
d) (u₁²f₂ – u₂²f₁) = f₁f₂(x)
View Answer

Answer: b
Explanation: By question the two motor cars approach each other along the same line.
Let P and Q be the positions of the cars on the line when each is seen from the other, where PQ = x.
It is evident that a collision can be just avoided, if the two cars stop at the point somewhere between P and Q that is if velocities of both the motor cars at O are zero.
Since the initial velocity of the motor car is at P is u₁ and it comes to rest at O, hence, its equation is
0 = u₁² – 2f₁(PO)
Or (PO) = u₁²/2f₁
Again, the initial velocity of the motor car at Q is due to and it comes to rest at O; hence its equation of motion is,
0 = u₂² – 2f₂(QO)
Or (QO) = u₂²/2f₂
Now, x = PQ = PO + OQ = u₁²/2f₁ + u₂²/2f₂
Or u₁²f₂ + u₂²f₁)/2f₁f₂ = x
Thus, (u₁²f₂ + u₂²f₁) = 2f₁f₂(x).

3. One motor car A stands 24m in front of a motorcycle B. Both starts from rest along a straight road in the same direction. If A moves with uniform acceleration of 2 m/sec² and B runs with a uniform velocity of 9 m/sec, is it possible for B to overtake A?
a) No
b) Yes
c) Data not sufficient
d) Answer cannot be determined
View Answer

Answer: a
Explanation: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.
If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.
Then, considering the motion of motor car A, we get
PR = 1/2 (2) (t²) = t²
Again, considering the motion of motorcycle B we get QR = 9t.
Now,QP+PR = QR
Or 24 + t² = 9t
Or t² – 9t + 24 = 0
Or t = [9 ± √(81 – 4*24)]/2
Or t = (9 ± √-15)/2
Clearly, the values of t are imaginary.
Therefore, there is no real positive time when B overtake A.
Hence, the in this case B will never overtake A.

4. One motor car A stands 24m in front of a motorcycle B. Both starts from rest along a straight road in the same direction. If A moves with uniform acceleration of 2 m/sec² and B runs at a uniform velocity of 11 m/sec then when will they meet?
a) After 3 seconds from start when the motorcycle B will overtake the motor car A
b) After 4 seconds from start when the motorcycle B will overtake the motor car A
c) After 2 seconds from start when the motorcycle B will overtake the motor car A
d) After 6 seconds from start when the motorcycle B will overtake the motor car A
View Answer

Answer: a
Explanation: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.
If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.
Then, considering the motion of motor car A, we get
PR = 1/2 (2) (t²) = t²
In this case when B runs ata uniform velocity of 11 m/sec, we shall have, QR=11t.
Therefore, in this case, QP + PR = QR gives,
Or 24 + t² = 11t
Or t² – 11t + 24 = 0
Or (t – 3)(t – 8) = 0
Or t = 3 or t = 8
Clearly, we are getting two real positive values of t.
Therefore, A and B will meet twice during the motion.
They will first meet after 3 seconds from start when the motorcycle B will overtake the motor car A.

5. One motor car A stands 24m in front of a motorcycle B. Both starts from rest along a straight road in the same direction. If A moves with uniform acceleration of 2 m/sec², B runs at a uniform velocity of 11 m/sec how many times will they meet?
a) 3
b) 2
c) 4
d) 1
View Answer

Answer: b
Explanation: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.
If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.
Then, considering the motion of motor car A, we get
PR = 1/2 (2) (t²) = t²
In this case when B runs at a uniform velocity 11 m/sec, we shall have, QR = 11t.
Therefore, in this case, QP + PR = QR gives,
Or 24 + t² = 11t
Or t² – 11t + 24 = 0
Or (t – 3)(t – 8) = 0
Or t = 3 or t = 8
Clearly, we are getting two real positive values of t.
Therefore, A and B will meet twice during the motion.

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6. One motor car A stands 24m in front of a motorcycle B. Both starts from rest along a straight road in the same direction. If A moves with uniform acceleration of 2 m/sec², then what will happen if B runs at a uniform velocity of 11 m/sec?
a) A will not overtake B
b) A will again overtake B and they will never meet again and again
c) A will again overtake B and they will meet again
d) A will again overtake B and they will never meet again
View Answer

Answer: d
Explanation: Let P and Q be the initial positions of the motor car and motorcycle B respectively, where PQ = 24m.
If possible, let us assume that B overtakes A after point R on the straight road after time t seconds from the start.
Then, considering the motion of motor car A, we get
PR = 1/2 (2) (t²) = t²
In this case when B runs at a uniform velocity 11 m/sec, we shall have, QR = 11t.
Therefore, in this case, QP + PR = QR gives,
Or 24 + t² = 11t
Or t² – 11t + 24 = 0
Or (t – 3)(t – 8) = 0
Or t = 3 or t = 8
Clearly, we are getting two real positive values of t.
Therefore, A and B will meet twice during the motion.
They will first meet after 3 seconds from start when the motorcycle B will overtake the motor car A.
But the velocity of A continuously increases, hence after a further period of (8 – 3) = 5 seconds A will again overtake B and they will never meet again.

7. Two particles start from rest from the same point and move along the same straight path; the first starts with a uniform velocity of 20 m/minute and the second with a uniform acceleration of 5 m/ min². Before they meet again, when will their distance be maximum?
a) At t = 8
b) At t = 6
c) At t = 4
d) At t = 2
View Answer

Answer: c
Explanation: Suppose, the two particle starts from rest at and move along the straight path OA.
Further assume that the distance between the particle is maximum after t minutes from start (before they meet again).
If we be the position of the particle after 30 minutes from the start which moves with uniform acceleration 5 m/min² and C that of the particle moving with uniform velocity 20 m/min, then we shall have,
OB = 1/2(5t²) and OC = 20t
If the distance between the particle after t minutes from start be x m, then,
x = BC = OC – OB = 20t – (5/2)t²
Now, dx/dt = 20 – 5t and d²x/dt² = -5
For maximum or minimum values of x, we have,
dx/dt = 0
Or 20 – 5t = 0
Or t = 4
And [d²y/dx²] = -5 < 0
Thus, x is maximum at t = 4.

8. Two particles start from rest from the same point and move along the same straight path; the first starts with a uniform velocity of 20 m/minute and the second with a uniform acceleration of 5 m/ min². Before they meet again, then what will be the maximum distance?
a) 38m
b) 36m
c) 42m
d) 40m
View Answer

Answer: d
Explanation: Suppose, the two particle starts from rest at and move along the straight path OA.
Further assume that the distance between the particle is maximum after t minutes from start (before they meet again).
If we be the position of the particle after 30 minutes from the start which moves with uniform acceleration 5 m/min² and C that of the particle moving with uniform velocity 20 m/min, then we shall have,
OB = 1/2(5t²) and OC = 20t
If the distance between the particle after t minutes from start be x m, then,
x = BC = OC – OB = 20t – (5/2)t² ……….(1)
Now, dx/dt = 20 – 5t and d²x/dt² = -5
For maximum or minimum values of x, we have,
dx/dt = 0
Or 20 – 5t = 0
Or t = 4
And [d²y/dx²] = -5 < 0
Thus, x is maximum at t = 4.
Therefore, the maximum value is,
= 20*4 – (5/2)(4²) [putting t = 4 in (1)]
= 40 m

9. An express train is running behind a goods train on the same line and in the same direction, their velocities being u₁ and u₂ (u₁ > u₂) respectively. When there is a distance x between them, each is seen from the other. At which point it is just possible to avoid a collision f₁ is the greatest retardation and f₂ is the greatest acceleration which can be produced in the two trains respectively?
a) (u₁ – u₂)² = 2x(f₁ + f₂)
b) (u₁ + u₂)² = 2x(f₁ – f₂)
c) (u₁ – u₂)² = 2x(f₁ – f₂)
d) (u₁ + u₂)² = 2x(f₁ + f₂)
View Answer

Answer: c
Explanation: Let A be the position of the express train and B, that of the goods train when each is seen from the other, where AB = x.
Evidently, to avoid a collision the express train will apply the greatest possible retardation f₁ and the goods train will apply the greatest possible acceleration f₂.
Now, it is just possible to avoid a collision,if the velocities of the two trains at a point C on the line are equal, when they just touch each other at C(because after the instant the velocity of the express train will decrease due to retardation f₁ and that of the goods train will increase due to acceleration f₂).
Let the two trains reach the point C, at time t after each is seen by the other and their equal velocities at C bev and BC = s.
Then the equation of motion of the express train is,
v = u₁ – f₁t ……….(1) And x + s = u₁t – (1/2)f₁t² ……….(2)
And the equation of the motion of the goods train is,
v = u₂ + f₂t ……….(3) And s = u₂t – (1/2)f₂t² ……….(4)
From (1) and (3) we get,
u₁ – f₁t = u₂ + f₂t
Or (f₁+ f₂)t = u₁ – u₂
Or t = (u₁ – u₂)/(f₁ + f₂)
Again, (2)–(4) gives,
x = (u₁ – u₂)t – ½(t²)(f₁ + f₂)
= t[(u₁ – u₂) – (1/2)(t)(f₁ + f₂)]
As, (f₁ + f₂)t = (u₁ – u₂)
So, x = (u₁ – u₂)/(f₁ + f₂)[(u₁ – u₂) – (1/2)(u₁ – u₂)]
So, (u₁ – u₂)² = 2x(f₁ – f₂)

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