# Mathematics Questions and Answers – Calculus Application – Motion in a Straight Line – 2

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This set of Mathematics Question Papers for Engineering Entrance Exams 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

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)

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

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

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

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.

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

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

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

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)

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) 