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NEET Questions

Q1:

Concept Videos :-

#42 Non Uniform Acceleration

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#43 Solved Examples 29

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Non-uniform acceleration

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elem

(1)

$t = \sqrt{x} + 3 x = {(t - 3)}^{2} A t t = 0 s e c, x_{1} = 9 m A t t = 6 s e c, x_{2} = 9 m D i s p l a c e m e n t = x_{2} - x_{1} = 0$

Difficulty Level:

(1)
60%
(2)
13%
(3)
14%
(4)
15%

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Q2:

Concept Videos :-

#42 Non Uniform Acceleration

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#43 Solved Examples 29

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#44 Solved Examples 30

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Concept Questions :-

Non-uniform acceleration

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elem

Difficulty Level:

(1)
22%
(2)
11%
(3)
65%
(4)
3%

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A point moves in a straight line under the retardation a $v^{2}$ . If the initial velocity is u, the distance covered in 't' seconds is-

1. $aut$

2. $\frac{1}{a} \ln (aut)$

3. $\frac{1}{a} \ln (1 + aut)$

4. $a \ln (aut)$

Q3:

Concept Videos :-

#42 Non Uniform Acceleration

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#43 Solved Examples 29

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#44 Solved Examples 30

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Non-uniform acceleration

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elem

Difficulty Level:

(1)
16%
(2)
24%
(3)
56%
(4)
6%

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A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by $(4 t^{3} - 2 t)$ , where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin ?

(1) 28 m/s²

(2) 22 m/s²

(3) 12 m/s²

(4) 10 m/s²

Q4:

Concept Videos :-

#42 Non Uniform Acceleration

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#43 Solved Examples 29

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#44 Solved Examples 30

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Non-uniform acceleration

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elem

(2) $v = 4 t^{3} - 2 t$ (given) ∴ $a = \frac{d v}{d t} = 12 t^{2} - 2$

and $x = \int_{0}^{t} v d t = \int_{0}^{t} (4 t^{3} - 2 t) d t = t^{4} - t^{2}$

When particle is at 2m from the origin $t^{4} - t^{2} = 2$

⇒ $t^{4} - t^{2} - 2 = 0$ ⇒ $(t^{2} - 2) (t^{2} + 1) = 0$ ⇒ $t = \sqrt{2} \sec$

Acceleration at $t = \sqrt{2} \sec$ given by,

$a = 12 t^{2} - 2 = 12 \times 2 - 2$ = $22 m / s^{2}$

Difficulty Level:

(1)
13%
(2)
67%
(3)
14%
(4)
9%

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The initial velocity of a particle is u (at t = 0) and the acceleration f is given by at. Which of the following relation is valid

(1) $v = u + a t^{2}$

(2) $v = u + a \frac{t^{2}}{2}$

(3) $v = u + a t$

(4) v = u

Q5:

PMT - 1981

Concept Videos :-

#42 Non Uniform Acceleration

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#43 Solved Examples 29

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#44 Solved Examples 30

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Non-uniform acceleration

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elem

(2) If acceleration is variable (depends on time) then

$a = \frac{d v}{d t} d v = a d t \int_{u}^{v} d v = \int_{0}^{t} (f t) d t$

$v = u + \frac{a t^{2}}{2}$

Difficulty Level:

(1)
27%
(2)
26%
(3)
31%
(4)
19%

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The velocity of a body depends on time according to the equation $v = 20 + 0 .1 t^{2}$ . The body is undergoing

(1) Uniform acceleration

(2) Uniform retardation

(3) Non-uniform acceleration

(4) Zero acceleration

Q6:

Concept Videos :-

#42 Non Uniform Acceleration

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#43 Solved Examples 29

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#44 Solved Examples 30

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Non-uniform acceleration

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elem

(3) Acceleration $a = \frac{d v}{d t} = 0 .1 \times 2 t = 0 .2 t$

Which is time dependent i.e. non-uniform acceleration.

Difficulty Level:

(1)
30%
(2)
8%
(3)
60%
(4)
5%

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A rocket is fired upward from the earth's surface such that it creates an acceleration of 19.6 m/sec². If after 5 sec its engine is switched off, the maximum height of the rocket from earth's surface would be [MP PET 1995]

(1) 245 m

(2) 490 m

(3) 980 m

(4) 735 m

Q7:

Concept Videos :-

#42 Non Uniform Acceleration

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#43 Solved Examples 29

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Non-uniform acceleration

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elem

(4) Given $a = 19 .6 m / s^{2} = 2 g$

Resultant velocity of the rocket after 5 sec

$v = 2 g \times 5 = 10 g m / s$

Height achieved after 5 sec, $h_{1} = \frac{1}{2} \times 2 g \times 25 = 245 m$

On switching off the engine it goes up to height $h_{2}$ where its velocity becomes zero.

$0 = {(10 g)}^{2} - 2 g h_{2} \Rightarrow h_{2} = 490 m$

∴ Total height of rocket $= 245 + 490 = 735 m$

Difficulty Level:

(1)
29%
(2)
22%
(3)
8%
(4)
42%

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A balloon starts rising from the ground with an acceleration of 1.25 m/s² . After 8s, a stone is released from the balloon. The stone will (g = 10 m/s²)

(1) Reach the ground in 4 second

(2) Begin to move down after being released

(3) Have a displacement of 50 m

(4) Cover a distance of 40 m in reaching the ground

Q8:

Concept Videos :-

#42 Non Uniform Acceleration

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#43 Solved Examples 29

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#44 Solved Examples 30

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Non-uniform acceleration

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elem

(1)

When the stone is released from the balloon. Its height $h = \frac{1}{2} a t^{2} = \frac{1}{2} \times 1 .25 \times {(8)}^{2} = 40 m$ and velocity

$v = a t = 1 .25 \times 8 = 10 m / s$

$T i m e t a k e n b y t h e s t o n e t o r e a c h t h e g r o u n d : h = u t + \frac{1}{2} a t^{2} - 40 = 10 t + \frac{1}{2} (- 10) t^{2} t^{2} - 2 t - 8 = 0 (t - 4) (t + 2) = 0 t = 4 s e c$

Difficulty Level:

(1)
50%
(2)
18%
(3)
15%
(4)
19%

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The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v₀ The distance travelled by the particle in time t will be

(1) $v_{0} t + \frac{1}{3} b t^{2}$

(2) $v_{0} t + \frac{1}{3} b t^{3}$

(3) $v_{0} t + \frac{1}{6} b t^{3}$

(4) $v_{0} t + \frac{1}{2} b t^{2}$

Q9:

PMT - 1995

Concept Videos :-

#42 Non Uniform Acceleration

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#43 Solved Examples 29

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#44 Solved Examples 30

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Non-uniform acceleration

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elem

(3) $\frac{d v}{d t} = b t \Rightarrow d v = b t d t \Rightarrow v = \frac{b t^{2}}{2} + K_{1}$

At $t = 0, v = v_{0} \Rightarrow K_{1} = v_{0}$

We get $v = \frac{1}{2} b t^{2} + v_{0}$

Again $\frac{d x}{d t} = \frac{1}{2} b t^{2} + v_{0} \Rightarrow x = \frac{1}{2} \frac{b t^{2}}{3} + v_{0} t + K_{2}$

At $t = 0, x = 0 \Rightarrow K_{2} = 0$

$∴ x = \frac{1}{6} b t^{3} + v_{0} t$

Difficulty Level:

(1)
6%
(2)
21%
(3)
49%
(4)
26%

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The acceleration of a particle starting from rest, varies with time according to the relation A = – aω² sinω t. The displacement of this particle at a time t will be

(1) $- \frac{1}{2} (a ω^{2} \sin ω t) t^{2}$