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/sec^{2}

b) 2 cm/sec^{2}

c) 3 cm/sec^{2}

d) 4 cm/sec^{2}

View Answer

Explanation: Let, the particle moving with a uniform acceleration of f cm/sec

^{2}.

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/sec

^{2}.

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

View Answer

Explanation: Let, the particle moving with a uniform acceleration of f cm/sec

^{2}.

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/sec

^{2}.

Thus, the space described by the particle in 8 seconds,

= [10*8 + 1/2(3)(8*8) [using the formula s = ut +1/2(ft

^{2})]

= 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 10^{th} second of its motion?

a) 38.5cm

b) 37.5cm

c) 38cm

d) 39.5cm

View Answer

Explanation: Let, the particle moving with a uniform acceleration of f cm/sec

^{2}.

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/sec

^{2}.

The space described during the 10th second of its motion is,

= [10 + 1/2(3)(2*10 – 1)] [using the formula s

_{t}= ut +1/2(f)(2t – 1)]

= 10 + 28.5

= 38.5cm.

4. A point starts with the velocity 10 cm/sec and moves along a straight line with uniform acceleration 5cm/sec^{2}. How much time it takes to describe 80 cm?

a) 4 seconds

b) 2 seconds

c) 8 seconds

d) 6 seconds

View Answer

Explanation: Let us assume that the point takes t seconds to describe a distance 80 cm.

Then using the formula s = ut +1/2(ft

^{2}) we get,

80 = 10*t + 1/2(5)(t

^{2})

Or 5t

^{2}+ 20t – 160 = 0

Or t

^{2}+ 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

View Answer

Explanation: Let f be the uniform retardation in m/sec

^{2}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(ft

^{2}) 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

View Answer

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/sec

^{2},

Then using the formula, v

^{2}= u

^{2}– 2fs, we get,

(u/2)

^{2}= u

^{2}– 2f(2.5)

Or f = 3u

^{2}/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 v

^{2}= u

^{2}– 2fs we get,

0 = u

^{2}– 2(3u

^{2}/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

View Answer

Explanation: Let the initial velocity of the particle be u cm/second and its uniform retardation be f cm/sec

^{2}.

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

s

_{t}= 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(ft

^{2}))

Putting u = ft, we get,

f/2 = 1/64((ft

^{2})/2)

Or t

^{2}= 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

View Answer

Explanation: Let the initial velocity of the particle be u cm/second and its uniform retardation be f cm/sec

^{2}.

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

s

_{t}= 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(ft

^{2}))

Putting u = ft, we get,

f/2 = 1/64((ft

^{2})/2)

Or t

^{2}= 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

View Answer

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 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, then what is the value of acceleration?

a) 2(bt_{1} – at_{2})/t_{1}t_{2}(t_{1} + t_{2})

b) -2(bt_{1} – at_{2})/t_{1}t_{2}(t_{1} + t_{2})

c) 2(bt_{1} + at_{2})/t_{1}t_{2}(t_{1} + t_{2})

d) 2(bt_{1} – at_{2})/t_{1}t_{2}(t_{1} – t_{2})

View Answer

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,

ut

_{1}+ ft

_{1}

^{2}/2 = a [as, AB = a] ……….(1)

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

u(t

_{1}+ t

_{2}) + f(t

_{1}+ t

_{2})

^{2}/2 = a + b [as, AC = AB + BC = a + b] ……….(2)

Multiplying (1) by (t

_{1}+ t

_{2}) and (2) by t

_{1}we get,

ut

_{1}(t

_{1}+ t

_{2}) + ft

_{1}

^{2}(t

_{1}+ t

_{2})/2 = a(t

_{1}+ t

_{2}) ……….(3)

And ut

_{1}(t

_{1}+ t

_{2}) + f(t

_{1}+ t

_{2})

^{2}/2 = (a + b)t

_{1}……….(4)

Subtracting (3) and (4) we get,

1/2(ft

_{1})(t

_{1}+ t

_{2})(t

_{1}– t

_{1}– t

_{2}) = at

_{2}– bt

_{1}

Solving the above equation, we get,

f = 2(bt

_{1}– at

_{2})/t

_{1}t

_{2}(t

_{1}+ t

_{2})

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