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Motion in a Straight Line MCQs for NEET — Physics Questions with Answers

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If the velocity of a particle is given by $v = {(180 - 16 x)}^{1 / 2}$ m/s, then its acceleration will be

Zero

8 m/s²

– 8 m/s²

4 m/s²

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Explanation

$v = {(180 - 16 x)}^{1 / 2}$

As $a = \frac{d v}{d t} = \frac{d v}{d x} . \frac{d x}{d t}$

$∴ a = \frac{1}{2} {(180 - 16 x)}^{- 1 / 2} \times (- 16) (\frac{d x}{d t})$

$= - 8 {(180 - 16 x)}^{- 1 / 2} \times v$

$= - 8 {(180 - 16 x)}^{- 1 / 2} \times {(180 - 16 x)}^{1 / 2}$ $= - 8 m / s^{2}$

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The displacement of a particle is proportional to the cube of time elapsed. How does the acceleration of the particle depends on time obtained

$a \propto t^{2}$

$a \propto t^{4}$

$a \propto t^{3}$

$a \propto t$

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Explanation

$x \propto t^{3}$ ∴ $x = K t^{3}$

$\Rightarrow v = \frac{d x}{d t} = 3 K t^{2}$ and $a = \frac{d v}{d t} = 6 K t$

i.e. $a \propto t$

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Speed of two identical cars are u and 4u at a specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is

1 : 1

1 : 4

1 : 8

1 : 16

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Explanation

$S \propto u^{2} \Rightarrow \frac{S_{1}}{S_{2}} = {(\frac{1}{4})}^{2} = \frac{1}{16}$

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A body is moving with uniform acceleration describes 40 m in the first 5 sec and 65 m in next 5 sec. Its initial velocity will be

4 m/s

2.5 m/s

5.5 m/s

11 m/s

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Explanation

For a body moving with uniform acceleration, the distance-time relation is given by: s = ut + (1/2)at^2. Using the given values of 40 m in 5 seconds and 65 m in the next 5 seconds, we can solve for the initial velocity (u) and acceleration (a). The solution yields an initial velocity of 5.5 m/s.

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The displacement x of a particle varies with time $t, x = a e^{- α t} + b e^{β t}$ , where $a, b, α and β$ are positive constants. The velocity of the particle will

Go on decreasing with time

Be independent of $α$ and $β$

Drop to zero when $α$ = $β$

Go on increasing with time

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Explanation

$x = a e^{- α t} + b e^{β t}$

Velocity $v = \frac{d x}{d t} = \frac{d}{d t} (a e^{- α t} + b e^{β t})$

$= a . e^{- α t} (- α) + b e^{β t} . β = - a α e^{- α t} + b β e^{β t}$

Acceleration $= - a α e^{- α t} (- α) + b β e^{b t} . β$

$= a α^{2} e^{- α t} + b β^{2} e^{β t}$

Acceleration is positive so velocity goes on increasing with time.

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A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate $\frac{f}{2}$ to come to rest. If the total distance traversed is 15 S, then

$S = \frac{1}{2} f t^{2}$

$S = \frac{1}{4} f t^{2}$

$S = \frac{1}{72} f t^{2}$

$S = \frac{1}{6} f t^{2}$

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A man is 45 m behind the bus when the bus starts accelerating from rest with acceleration of 2.5 m/s². With what minimum velocity should the man start running to catch the bus?

12 m/s

14 m/s

15 m/s

16 m/s

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Explanation

Let man will catch the bus after 't' sec . So he will cover distance ut.

Similarly distance travelled by the bus will be $\frac{1}{2} a t^{2}$ . For the given condition

$u t = 45 + \frac{1}{2} a t^{2} = 45 + 1 .25 t^{2}$ $[A s a = 2 .5 m / s^{2}]$

⇒ $u = \frac{45}{t} + 1 .25 t$

To find the minimum value of u

$\frac{d u}{d t} = 0$

so we get $t = 6 \sec$ then,

$u = \frac{45}{6} + 1 .25 \times 6 = 7 .5 + 7 .5 = 15 m / s$

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A particle moves along x-axis as $x = 4 (t - 2) + a {(t - 2)}^{2}$ . Which of the following is true ?

The initial velocity of particle is 4

The acceleration of particle is 2a

The particle is at origin at t = 0

None of these

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Explanation

$x = 4 (t - 2) + a {(t - 2)}^{2}$

At $t = 0, x = - 8 + 4 a = 4 a - 8$

$v = \frac{d x}{d t} = 4 + 2 a (t - 2)$

At $t = 0, v = 4 - 4 a = 4 (1 - a)$

But acceleration, $a = \frac{d^{2} x}{d t^{2}} = 2 a$

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A body starting from rest moves with constant acceleration. The ratio of distance covered by the body during the 5th sec to that covered in 5 sec is

9/25

3/5

25/9

1/25

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Explanation

Distance covered in 5^th second,

$S_{5^{t h}} = u + \frac{a}{2} (2 n - 1) = 0 + \frac{a}{2} (2 \times 5 - 1) = \frac{9 a}{2}$

and distance covered in 5 second,

$S_{5} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} \times a \times 25 = \frac{25 a}{2}$

∴ $\frac{S_{5^{t h}}}{S_{5}} = \frac{9}{25}$

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Two trains, each 50 m long are travelling in opposite direction with velocity 10 m/s and 15 m/s. The time of crossing is

10 s

4 s

$2 \sqrt{3} s$

$4 \sqrt{3} s$

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Explanation

Time $= \frac{Total length}{Relative velocity} = \frac{50 + 50}{10 + 15} = \frac{100}{25} = 4 s e c$

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