NEET Physics Motion in A Straight Line Questions Solved

A ball is thrown upward with a certain speed. It passes through the same point at 3 second and 7 second from the start. The maximum height achieved by the ball is

1. 500 m

2. 250 m

3. 125 m

4. 450 m

#Uniformly accelerated motion

(3)

As upward and downward motion are symmetric [Time taken from A $\to$ B is same as time taken from B $\to$ C]

So time to reach highest point is 5s. Same time is required to reach from B to ground.

Time to reach highest point after 3sec = $\frac{7 - 3}{2} = 2 s e c .$

Time to reach highest point from ground = 3+2 = 5sec.

For downward motion,

$u = 0 a = g t = 5 s h = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times 5^{2} = 125 m$

Difficulty Level:

(1)
17%
(2)
29%
(3)
50%
(4)
6%

#Uniformly accelerated motion

$a = \frac{d v}{d t} = \frac{d v}{d s} \times \frac{d s}{d t} = v \times \frac{d v}{d s}$

Difficulty Level:

(1)
44%
(2)
22%
(3)
22%
(4)
14%

#Relative motion in 1-D

(1)

$Motion of A wrt B$

$Relative velocity of A wrt B = \vec{V_{A}} - \vec{V_{B}} = v - (vcos 60 °) = v / 2$

$Time = \frac{a}{\frac{v}{2}} = \frac{2 a}{v}$

Difficulty Level:

(1)
37%
(2)
32%
(3)
23%
(4)
10%

#Uniformly accelerated motion

Distance travelled in t second,

$H e r e, i n i t i a l v e l o c i t y u = 0 h = \frac{g t^{2}}{2} . . . . . . (1) D i s \tan c e t r a v e l l e d i n n^{t h} s e c o n d = u + \frac{a}{2} (2 n - 1) D i s \tan c e t r a v e l l e d i n l a s t s e c o n d = 0 + \frac{g}{2} (2 t - 1) = \frac{9 h}{25} . . . . . . . . . . . . . . (2) E q u a t i n g h f r o m e q u a t i o n 1 a n d 2 \frac{25 g (2 t - 1)}{18} = \frac{g t^{2}}{2} 9 t^{2} - 50 t + 25 = 0 S o l v i n g f o r t, t = \frac{50 \pm \sqrt{50^{2} - 4 \times 9 \times 25}}{2 \times 9} = \frac{10}{18} s o r 5 s T i m e h a s t o b e g r e a t e r t h a n 1 s e c o n d, s o t = 5 s P u t t i n g v a l u e o f t i n e q n . 1 h = \frac{g {(5)}^{2}}{2} = 122.5 m$

Difficulty Level:

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16%
(2)
22%
(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. $a \ln (aut)$

#Non-uniform acceleration

Difficulty Level:

(1)
17%
(2)
30%
(3)
47%
(4)
8%

#Uniformly accelerated motion

(2)

Upward journey Downward journey

Net acceleration = -(a+g) Net acceleration = g - a
-ve indicates retardation

$t_{v} = \sqrt{\frac{2 h}{g + a}} t_{D} = \sqrt{\frac{2 h}{g - a}} \frac{t_{U}}{T_{D}} = \sqrt{\frac{g - a}{g + a}} = \sqrt{\frac{8}{12}} = \sqrt{\frac{2}{3}}$

Difficulty Level:

(1)
39%
(2)
34%
(3)
17%
(4)
11%

#Relative motion in 1-D

Difficulty Level:

(1)
14%
(2)
50%
(3)
28%
(4)
9%