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__AIEEE CornerSubjective Questions (Level 1) __

**Basic Definitions****Ques 1: A car moves with 60 km/h in first one hour and with 80 km/h in next half an hour. Find: ****(a) total distance travelled by the car, ****(b) average speed of car in total 1.5 hours.****Ans: **(a) 100 km

(b) 66.67 kmh^{-1}**Sol: **(a) D = v_{1}t_{1} + v_{2} t_{2}

(b) Average speed**Ques 2: A particle moves in a straight line with initial velocity 4 m/s and a constant acceleration of 6 m/s ^{2}. Find the average velocity of the particle in a time interval from **

(b) 22 ms

= 20 m

âˆ´ Average velocity = 20m/2s = 10m/s

(b) Displacement in first four seconds

âˆ´ Displacement in the time interval

t = 2s to t = 4s

= 64 - 20

= 44 m

âˆ´ Average velocity = 44m/2s = 22 m/s.

(b) 10 ms

âˆ´

i.e., 5t

t = 6.1 s

If the particle goes h meter above tower before coming down

0 = (20)

â‡’ h = 20 m

= 9.8 m/s (downwards)

âˆ´ 5t

or T = 4t

(b) 8 ms

(c) 8 ms

(b) As the particle did not return back distance travelled in 12 s

= Displacement at 12 s

âˆ´ Average speed = 8 m/s.

âˆ´ (âˆµ a = 4 m/s

i.e., v

Displacement of particle in 12 seconds

= Area under v-t graph

Average velocity

= +8 m/s

âˆ´ Circumference of circle = 2Ï€R

Speed (v) of particle = 1 m/s

âˆ´ Distance moved by particle in 2 s = 2 m

Thus, angle through which the particle moved

Magnitude of Average velocity

(b) Magnitude of average acceleration

Position vector at t = 4 s

(a) Displacement from t = 0 s to t = 4 s

(b) Average acceleration

(c) We cannot find the average speed as the actual path followed by the particle is not known.

**Uniform acceleration(a) One dimensional motion**

**Ques 8: Two diamonds begin a free fall from rest from the same height, 1.0 s apart. How long after the first diamond begins to fall will the two diamonds be 10 m apart? Take g = 10 m/s ^{2}.**

or t

or t

or 2t = 3

t = 1.5 s

Displacement of first after attaining highest point = Displacement of second before attaining highest point

or

or

or

**Sol: **

â‡’ t = 1s

For A

or ** ** â€¦(i)

For B

or

[Substituting value of H from Eq. (i)]

2t - 1 = 5

â‡’ t = 3 s

Substituting t = 3 s in Eq. (i)**Ques 11: A point mass starts moving in a straight line with constant acceleration. After time t _{0} the acceleration changes its sign, remaining the same in magnitude. Determine the time t from the beginning of motion in which the point mass returns to the initial position.**

âˆ´

or

or

(- ive sign being absurd)

From the begining of the motion the point mass will return to the initial position after time 3.141t

â‡’ u

(a) For H

0

i.e., 20H = 325

or H = 16.25 m

(b) For t

âˆ´

= 1.8 s

(b)1 .6 7 ms

(c) 7.5 m

(a) 15

and 15 = u + a x 6 â€¦(ii)

Substituting the value of 6a from Eq. (ii) in Eq. (i)

225 = u

i.e., u

(u - 15)(u - 5) = 0

âˆ´ u = 5 m/s

(15 m/s being not possible)

(b) Using Eq. (ii)

(c) u

i.e.,

= 7.5 m

(d)

â€¦(iii)

Differentiating Eq. (iii) w.r.t. time t

Journey A to P

v

and â€¦(ii)

Journey P to B

0 = v

and â€¦(iv)

â‡’

âˆ´

or â€¦(v)

From Eq. (ii) and Eq. (iv)

or â€¦(vi)

Dividing Eq. (vi) by Eq. (v)

v

Substituting the value of v

(Proved)

v = u + at

0 = u + a6

âˆ´

(a) At t = 10 s, s = - 2 m

or -2 = (-6a) 10 + 50a

or -10a = - 2

or a = 0.2 m/s

(b) v(at t = 10 s) = u + a 10

= - 6a + 10a

= 4 a

= 0.8 m/s

**(b) Two or three dimensional motion**

**Ques 16: Net force acting on a particle of mass 2 kg is 10 N in north direction. At t = 0, particle was moving eastwards with 10 m/s. Find displacement and velocity of particle after 2 s.****Ans: **10âˆš5 m at cos^{-1}(2) from east towards north, 10âˆš2 ms-1 at 45Â° from east towards north.**Sol:**

m at cot^{-1}(2) from east to north.**Ques 17: At time t = 0, a particle is at (2m, 4m). It starts moving towards positive x-axis with constant acceleration 2 m/s ^{2} (initial velocity = 0). After 2 s an acceleration of 4 m/s^{2} starts acting on the particle in negative y-direction also. Find after next 2 s: **

(a) Velocity

or

(b) Co-ordinate of particle

Co-ordinate of the particle [18 m, - 4 m]**Ques 18: A particle moving in x-y plane is at origin at time t = 0. Velocity of the particle is and acceleration is ****(a) velocity of particle and (b) coordinates of particle.**

(b) Co-ordinates of the particle

âˆ´ Co-ordinates of particle would be [10 m , - 2 m]

Comparing the coefficients of

29 = 2t

and n = 8t + t

â‡’

Substituting value of t in Eq. (ii)

n = 8 x 3.807 + (3.807)

= 44.95

= 24.44 m/s

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