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 t_{1} = time taken to go from A to B; t_{2} = 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 p^{2}, q^{2} and r^{2}?

a) A.P

b) G.P

c) H.P

d) They are in any arbitrary series

View Answer

Explanation: When a = b, then the equation of the motion of the particle from A to B is,

q

^{2}= p

^{2}+ 2fa

Or q

^{2}– p

^{2}= 2fa ……….(1)

Again, the equation of motion of the particle from B to C is,

r

^{2}= q

^{2}+ 2fa

Or r

^{2}– q

^{2}= 2fa ……….(2)

From (1) and (2) we get,

q

^{2}– p

^{2}= r

^{2}– q

^{2}

Thus, p

^{2}, q

^{2}and r

^{2}are in A.P.

2. Two motor cars on the same line approach each other with velocities u_{1} and u_{2} respectively. When each is seen from the other, the distance between them is x. If f_{1} and f_{2} 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) (u_{1}^{2}f_{2} – u_{2}^{2}f_{1}) = 2f_{1}f_{2}(x)

b) (u_{1}^{2}f_{2} + u_{2}^{2}f_{1}) = 2f_{1}f_{2}(x)

c) (u_{1}^{2}f_{2} + u_{2}^{2}f_{1}) = f_{1}f_{2}(x)

d) (u_{1}^{2}f_{2} – u_{2}^{2}f_{1}) = f_{1}f_{2}(x)

View Answer

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 u

_{1}and it comes to rest at O, hence, its equation is

0 = u

_{1}

^{2}– 2f

_{1}(PO)

Or (PO) = u

_{1}

^{2}/2f

_{1}

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 = u

_{2}

^{2}– 2f

_{2}(QO)

Or (QO) = u

_{2}

^{2}/2f

_{2}

Now, x = PQ = PO + OQ = u

_{1}

^{2}/2f

_{1}+ u

_{2}

^{2}/2f

_{2}

Or u

_{1}

^{2}f

_{2}+ u

_{2}

^{2}f

_{1})/2f

_{1}f

_{2}= x

Thus, (u

_{1}

^{2}f

_{2}+ u

_{2}

^{2}f

_{1}) = 2f

_{1}f

_{2}(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/sec^{2} 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

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) (t

^{2}) = t

^{2}

Again, considering the motion of motorcycle B we get QR = 9t.

Now,QP+PR = QR

Or 24 + t

^{2}= 9t

Or t

^{2}– 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/sec^{2} 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

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) (t

^{2}) = t

^{2}

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 + t

^{2}= 11t

Or t

^{2}– 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/sec^{2}, 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

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) (t

^{2}) = t

^{2}

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 + t

^{2}= 11t

Or t

^{2}– 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.

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/sec^{2}, 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

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) (t

^{2}) = t

^{2}

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 + t

^{2}= 11t

Or t

^{2}– 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/ min^{2}. 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

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/min

^{2}and C that of the particle moving with uniform velocity 20 m/min, then we shall have,

OB = 1/2(5t

^{2}) 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)t

^{2}

Now, dx/dt = 20 – 5t and d

^{2}x/dt

^{2}= -5

For maximum or minimum values of x, we have,

dx/dt = 0

Or 20 – 5t = 0

Or t = 4

And [d

^{2}y/dx

^{2}] = -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/ min^{2}. Before they meet again, then what will be the maximum distance?

a) 38m

b) 36m

c) 42m

d) 40m

View Answer

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/min

^{2}and C that of the particle moving with uniform velocity 20 m/min, then we shall have,

OB = 1/2(5t

^{2}) 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)t

^{2}……….(1)

Now, dx/dt = 20 – 5t and d

^{2}x/dt

^{2}= -5

For maximum or minimum values of x, we have,

dx/dt = 0

Or 20 – 5t = 0

Or t = 4

And [d

^{2}y/dx

^{2}] = -5 < 0

Thus, x is maximum at t = 4.

Therefore, the maximum value is,

= 20*4 – (5/2)(4

^{2}) [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 u_{1} and u_{2} (u_{1} > u_{2}) 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 f_{1} is the greatest retardation and f_{2} is the greatest acceleration which can be produced in the two trains respectively?

a) (u_{1} – u_{2})^{2} = 2x(f_{1} + f_{2})

b) (u_{1} + u_{2})^{2} = 2x(f_{1} – f_{2})

c) (u_{1} – u_{2})^{2} = 2x(f_{1} – f_{2})

d) (u_{1} + u_{2})^{2} = 2x(f_{1} + f_{2})

View Answer

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 f

_{1}and the goods train will apply the greatest possible acceleration f

_{2}.

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 f

_{1}and that of the goods train will increase due to acceleration f

_{2}).

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 = u

_{1}– f

_{1}t ……….(1) And x + s = u

_{1}t – (1/2)f

_{1}t

^{2}……….(2)

And the equation of the motion of the goods train is,

v = u

_{2}+ f

_{2}t ……….(3) And s = u

_{2}t – (1/2)f

_{2}t

^{2}……….(4)

From (1) and (3) we get,

u

_{1}– f

_{1}t = u

_{2}+ f

_{2}t

Or (f

_{1}+ f

_{2})t = u

_{1}– u

_{2}

Or t = (u

_{1}– u

_{2})/(f

_{1}+ f

_{2})

Again, (2)–(4) gives,

x = (u

_{1}– u

_{2})t – ½(t

^{2})(f

_{1}+ f

_{2})

= t[(u

_{1}– u

_{2}) – (1/2)(t)(f

_{1}+ f

_{2})]

As, (f

_{1}+ f

_{2})t = (u

_{1}– u

_{2})

So, x = (u

_{1}– u

_{2})/(f

_{1}+ f

_{2})[(u

_{1}– u

_{2}) – (1/2)(u

_{1}– u

_{2})]

So, (u

_{1}– u

_{2})

^{2}= 2x(f

_{1}– f

_{2})

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