Mathematics Problems for Engineering Entrance Exams on “Calculus Application – Velocity – 2”.
1. A particle moves in a straight-line OA; the distance of the particle from O at time t seconds is x ft, where x = a + bt + ct2 (a, b > 0). What will be the nature of motion of the particle when c = 0?
a) Uniform retardation
b) Uniform speed
c) Uniform acceleration
d) Uniform velocity
Answer: d
Clarification: We have, x = a + bt + ct2 ……….(1)
Let, v and f be the velocity and acceleration of a particle at time t seconds.
Then, v = dx/dt = d(a + bt + ct2)/dt = b + ct ……….(2)
And f = dv/dt = d(b + ct)/dt = c ……….(3)
Clearly, when c = 0, then f = 0 that is, acceleration of the particle is zero.
Hence in this case the particle moves with an uniform velocity.
2. A particle moves in a straight line and its velocity v at time t seconds is given byv = 3t2 – 4t + 5 cm/second. What will be the distance travelled by the particle during first 3 seconds after the start?
a) 21 cm
b) 22 cm
c) 23 cm
d) 24 cm
Answer: d
Clarification: Let, x be the distance travelled by the particle in time t seconds.
Then, v = dx/dt = 3t2 – 4t + 5
Or ∫dx = ∫ (3t2 – 4t + 5)dt
So, on integrating the above equation, we get,
x = t3 – 2t2 + 5t + c where, c is a constant. ……….(1)
Therefore, the distance travelled by the particle at the end of 3 seconds,
= [x]t = 3 – [x]t = 0
= (33 – 2*32 + 5*3 + c) – c [using (1)]
= 24 cm.
3. A particle moving in a straight line traverses a distance x in time t. If t = x2/2 + x, then which one is correct?
a) The retardation of the particle is the cube of its velocity
b) The acceleration of the particle is the cube of its velocity
c) The retardation of the particle is the square of its velocity
d) The acceleration of the particle is the square of its velocity
Answer: a
Clarification: We have, t = x2/2 + x
Therefore, dt/dx = 2x/2 + 1 = x + 1
Thus, if v be the velocity of the particle at time t, then
v = dx/dt = 1/(dt/dx)
= 1/(x + 1) = (x + 1)-1
Thus dv/dt = d((x + 1)-1)/dt
= (-1)(x + 1)-2 d(x + 1)/dt
= -1/(x + 1)2 * dx/dt
As, 1/(x + 1) = dx/dt,
So, -(dx/dt)2(dx/dt)
Or dv/dt = -v2*v [as, dx/dt = v]
= -v3
We know, dv/dt = acceleration of a particle.
As, dv/dt is negative, so there is a retardation of the particle.
Thus, the retardation of the particle = -dv/dt = v3 = cube of the particle.
4. A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds,respectively.What is the distance described by the particle in 3 seconds?
a) 30 cm
b) 31 cm
c) 32 cm
d) 33 cm
Answer: b
Clarification: We assume that the particle moves with uniform acceleration 2f m/sec.
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft2 + bt + a ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21 ……….(3)
16f + 4b + a = 43 ……….(4)
49f + 7b + a = 91 ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t2 + 5t + 7
Therefore, the distance described by the particle in 3 seconds,
= [x]t = 3 = (32 + 5*3 + 7)m = 31m
5. A particle moves with uniform acceleration along a straight line and describes distances 21m, 43m and 91m at times 2, 4 and 7 seconds, respectively.What is the velocity of the particle in 3 seconds?
a) 11m/sec
b) 31 cm/sec
c) 21m/sec
d) 41m/sec