The document Motion in One Dimension: JEE Main(Part- 6) - Physics, Solution by DC Pandey NEET Notes | EduRev is a part of the NEET Course DC Pandey (Questions & Solutions) of Physics: NEET.

All you need of NEET at this link: NEET

**Ques 21: Two objects are moving along the same straight line. They cross a point A with an acceleration a, 2a and velocity 2u, u at time t = 0. The distance moved by the object when one overtakes the other is****Ans:** **Sol:**

⇒

or

or t = 2u/a

∴

Option (a) is correct.**Ques 22: A cart is moving horizontally along a straight line with constant speed 30 ms ^{-1}. A particle is to be fired vertically upwards from the moving cart in such a way that it returns to the cart at the same point from where it was projected after the cart has moved 80 m. At what speed (relative to the cart) must the projectile be fired? (Take g = 10ms^{-2})**

i.e., 2u/10 = 8/3

or

Option (c) is correct.

∴ Displacement of particle at time (t = 2)

= (+ 5 m) + (-5 m) = 0 m.

i.e., the particle crosses its initial position at t = 2 s.

Option (b) is correct.

Thus, option (a) and (c) are incorrect.

Now, as the ball will have +ive acceleration throughout its motion option (b) is also incorrect.

∴ Correct option is (d).

Displacement of ball in time t = Displacement of lift in time t

or

∴

Option (a) is correct.

∴

or

or

or

Now, at t = 0, v = 10 m/s (given)

∴

i.e.,

∵ For velocity v at time t (= 2 s)

⇒ v = + 2 m/s.

Option (a) is correct.

0

i.e., S

For displacement (S

0

i.e., S

∴ S

= 225 m

Option (b) is correct

∴ displacement (S) of ball 1 at time t

= displacement (S) of ball 2 at time (t - 2)

or

or (t + t - 2) (t - t + 2) = 16

or 2t - 2 = 8

or t = 5 s

= 200 - 125

= 75 m

Option (b) is correct.

i.e., u = gT …(i)

…(ii)

and …(iii)

Substracting Eq. (ii) from Eq. (iii),

Option (d) is correct.

i.e., x = t

i.e., v ∝ t

Option (b) is correct.

= Area under curve (t-a graph)

= 90 m/s

At s = 0 (i.e., initially),

velocity of particle = 5 m/s

Option (b) is correct.

Differentiating equation (i) w.r.t. s

Option (b) is correct.

Option (b) is correct.

⇒ θ = 30°

Option (b) is correct.

i.e., ma

or

or a

or

or

or

or v = 0.1t

Now, at t = 0, v = 10 m/s

∴ 10 = k

Thus, v = 0.1 t

∴ at t = 5 s

v = 0.1 (5)

= 12.5 - 16.0 + 10

= 6.5 m/s Option (b) is correct.

⇒

⇒ u

Thus,

or gH = 100

⇒ H = 10 m

Option (b) is correct.

102 = u^{2} - 2g x 15

100 = v^{2} - 2 x 10 x 15

i.e., u = 20 m/s

Now, v = u - gt = 20 - 10 x 3

= 20 - 10 × (3)

= - 10 m/s, downward

Option (d) is correct.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!