# Mathematics Questions and Answers – Calculus Application – Motion in a Straight Line – 1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Motion in a Straight Line – 1”.

1. A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the acceleration?
a) 1 cm/sec2
b) 2 cm/sec2
c) 3 cm/sec2
d) 4 cm/sec2

Explanation: Let, the particle moving with a uniform acceleration of f cm/sec2.
By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.
Thus, using the formula v = u + ft we get
34 = 10 + f*8
Or 8f = 24
Or f = 3
Therefore, the required acceleration of the particle is 3 cm/sec2.

2. A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the value of space described?
a) 172 cm
b) 176 cm
c) 178 cm
d) 174 cm

Explanation: Let, the particle moving with a uniform acceleration of f cm/sec2.
By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.
Thus, using the formula v = u + ft we get
34 = 10 + f*8
Or 8f = 24
Or f = 3
Therefore, the required acceleration of the particle is 3 cm/sec2.
Thus, the space described by the particle in 8 seconds,
= [10*8 + 1/2(3)(8*8) [using the formula s = ut +1/2(ft2)]
= 80 + 96
= 176 cm.

3. A particle moving in a straight line with uniform acceleration has a velocity 10 cm/sec and 8 seconds later has a velocity 54 cm/sec. What will be the described by the particle during the 10th second of its motion?
a) 38.5cm
b) 37.5cm
c) 38cm
d) 39.5cm

Explanation: Let, the particle moving with a uniform acceleration of f cm/sec2.
By question initial velocity of the particle is u = 10 cm/sec and its velocity after 8 seconds = v = 34 cm/sec.
Thus, using the formula v = u + ft we get
34 = 10 + f*8
Or 8f = 24
Or f = 3
Therefore, the required acceleration of the particle is 3 cm/sec2.
The space described during the 10th second of its motion is,
= [10 + 1/2(3)(2*10 – 1)]   [using the formula st = ut +1/2(f)(2t – 1)]
= 10 + 28.5
= 38.5cm.
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4. A point starts with the velocity 10 cm/sec and moves along a straight line with uniform acceleration 5cm/sec2. How much time it takes to describe 80 cm?
a) 4 seconds
b) 2 seconds
c) 8 seconds
d) 6 seconds

Explanation: Let us assume that the point takes t seconds to describe a distance 80 cm.
Then using the formula s = ut +1/2(ft2) we get,
80 = 10*t + 1/2(5)(t2)
Or 5t2 + 20t – 160 = 0
Or t2 + 4t – 32 = 0
Or (t – 4)(t + 8) = 0
Or t = 4 Or -8
Clearly, t = -8 is inadmissible.
Therefore, the required time = 4 seconds.

5. A motor car travelling at the rate of 40 km/hr is stopped by its brakes in 4 seconds. How long will it go from the point at which the brakes are first applied?
a) 22m
b) 22(2/9)m
c) 22(1/9)m
d) 22(4/9)m

Explanation: Let f be the uniform retardation in m/sec2 to the motion of the motor car due to application of brakes.
By question, the car is stopped by its brakes in 4 seconds, hence, the final velocity of the car after 4 seconds = 0.
Therefore, using the formula v = u – ft we get,
0 = ((40*1000)/(60*60) – f(4)) [Since u = initial velocity of the motor car = 40 km/hr = (40*1000)/(60*60) m/sec]
Or f = 25/9
Let the car go through a distance s m from the point at which the brakes are first applied.
Then using the formula s = ut – 1/2(ft2) we get,
s = ((40*1000)/(60*60))*4 – 1/2(25/9)(4*4)
= 200/9
= 22(2/9)
Therefore, the required distance described by the car = 22(2/9)m.

6. A bullet fired into a target loses half of its velocity after penetrating 2.5 cm. How much further will it penetrate?
a) 0.85 cm
b) 0.84 cm
c) 0.83 cm
d) 0.82 cm

Explanation: Let, u cm/sec be the initial velocity of the bullet.
By the question, the velocity of the bullet after pertaining 2.5 cm into the target will be u/2 cm/sec.
Now if the uniform retardation to penetration be f cm/sec2,
Then using the formula, v2 = u2 – 2fs, we get,
(u/2)2 = u2 – 2f(2.5)
Or f = 3u2/20
Now let us assume that the bullet can penetrate x cm into the target.
Then the final velocity of the bullet will be zero after penetrating x cm into the target.
Hence, using formula v2 = u2 – 2fs we get,
0 = u2 – 2(3u2/20)(x)
Or 10 – 3x = 0
Or x = 10/3 = 3.33(approx).
Therefore, the required further penetration into the target will be
3.33 – 2.5 = 0.83 cm.

7. A particle moving in a straight line with uniform retardation described 7cm in 5th second and after some time comes to rest. If the particle describes 1/64 part of the total path during the last second of its motion, for how long was the particle in motion?
a) 6 seconds
b) 8 seconds
c) 4 seconds
d) 2 seconds

Explanation: Let the initial velocity of the particle be u cm/second and its uniform retardation be f cm/sec2.
Further assume that the particle was in motion for t seconds.
By question, the particle comes to rest after t seconds.
Therefore, using the formula, v = u – ft, we get,
0 = u – ft
Or u = ft
Again, the particle described 7cm in the 5th second. Therefore, using the formula
st = ut + 1/2(f)(2t – 1) we get,
7 = u – ½(f)(2.5 – 1)
Or u – 9f/2 = 7
Again, the distance described in the last second (i.e., in the t th second) of its motion
= 1/64 * (distance described by the particle in t seconds)
ut -1/2(f)(2t – 1) = 1/64(ut – ½(ft2))
Putting u = ft, we get,
f/2 = 1/64((ft2)/2)
Or t2 = 64
Or t = 8 seconds.

8. A particle moving in a straight line with uniform retardation described 7cm in 5th second and after some time comes to rest. If the particle describes 1/64 part of the total path during the last second of its motion, for how long was the particle in motion?
a) 10 cm/sec
b) 12 cm/sec
c) 14 cm/sec
d) 16 cm/sec

Explanation: Let the initial velocity of the particle be u cm/second and its uniform retardation be f cm/sec2.
Further assume that the particle was in motion for t seconds.
By question, the particle comes to rest after t seconds.
Therefore, using the formula, v = u – ft, we get,
0 = u – ft
Or u = ft ……….(1)
Again, the particle described 7cm in the 5th second. Therefore, using the formula
st = ut +1/2(f)(2t – 1) we get,
7 = u – ½(f)(2.5 – 1)
Or u – 9f/2 = 7 ……….(2)
Again, the distance described in the last second (i.e., in the t th second) of its motion
= 1/64 * (distance described by the particle in t seconds)
ut – 1/2(f)(2t – 1) = 1/64(ut – ½(ft2))
Putting u = ft, we get,
f/2 = 1/64((ft2)/2)
Or t2 = 64
Or t = 8 seconds.
Therefore, from (1) we get, u = 8f
Putting u = 8f in (2) we get,
8f – 9f/2 = 7
Or f = 2
Thus, u = 8f = 8*2 = 16
Therefore, the particle was in motion for 8 seconds and its initial velocity is 16 cm/sec.

9. If a, b, c be the space described in the pth, qth and rth seconds by a particle with a given velocity and moving with uniform acceleration in a straight line then what is the value of a(q – r) + b(r – p) + c(p – q)?
a) 0
b) 1
c) -1
d) Can’t be determined

Explanation: Let, f be the uniform acceleration and u be the given initial velocity of the moving particle.
From the conditions of problem we have the following equation of motion of the particle:
u + ½(f)(2p – 1) = a ……….(1)
u + ½(f)(2q – 1) = b ……….(2)
u + ½(f)(2r – 1) = c ……….(1)
Thus, a(q – r) + b(r – p) + c(p – q) = [u + ½(f)(2p – 1)](q – r) + [u + ½(f)(2q – 1)](r – p) + [u + ½(f)(2r – 1)](p – q)
= u (q – r + r – p + p – q) + f/2[(2p – 1)(q – r) + (2q – 1)(r – p) + (2r – 1)(p – q)] = u*0 + f/2[2(pq – rp + qr – pq + rp – qr) – q + r – r + p – p + q]
= f/2*0
= 0

10. 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, then what is the value of acceleration?
a) 2(bt1 – at2)/t1t2(t1 + t2)
b) -2(bt1 – at2)/t1t2(t1 + t2)
c) 2(bt1 + at2)/t1t2(t1 + t2)
d) 2(bt1 – at2)/t1t2(t1 – t2)

Explanation: Let the particle beam moving with uniform acceleration f and its velocity at A be u.
Then, the equation of the motion of the particle from A to B is,
ut1 + ft12/2 = a   [as, AB = a] ……….(1)
Again, the equation of motion of the particle from A to C is,
u(t1 + t2) + f(t1 + t2)2/2 = a + b  [as, AC = AB + BC = a + b] ……….(2)
Multiplying (1) by (t1 + t2) and (2) by t1 we get,
ut1 (t1 + t2) + ft12(t1 + t2)/2 = a(t1 + t2) ……….(3)
And ut1(t1 + t2) + f(t1 + t2)2/2 = (a + b)t1 ……….(4)
Subtracting (3) and (4) we get,
1/2(ft1)(t1 + t2)(t1 – t1 – t2) = at2 – bt1
Solving the above equation, we get,
f = 2(bt1 – at2)/t1t2(t1 + t2)

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