Test: Motion in a Plane- 2 - NEET MCQ

Initial velocity = u = 20 m/s 
Final velocity = v = 0     ( at maximum height , v = 0 )
Acceleration due to gravity(g)  in this case, is taken as negative.
This is because when the direction of motion of object is opposite to "g" , then value of g is taken as -ve.
Hence, g = -9.8 m/s²
Using v² - u² = 2gh
0² - 20² = 2 * (- 9.8) * h
- 400 = -19.6 h
h = - 400 /(-19.6) = 20.408 m (approximately)

Use equation of kinematics: v = u + at
Initial velocity, u = 3i + 4j
Acceleration, a = 0.4i + 0.3j
Time, t = 10s
v = (3i + 4j) + 10(0.4i + 0.3j)
v = (3i + 4j) + (4i + 3j)
v = 7i + 7j
|v| = √(7² + 7²) = 7√2

A body start from rest so, u = 0 
Let total distance covered = s
Let a body moves with accerlation = a 
s = 1/2 × a× t²
s = 1/2× a × 4²
s = 8a
The one-fourth of total distance, s'  = 8a/4
= 2a
s' = 1/2 × a × t²
2a / a = 1/2 × t²
4 = t²
t = 2 seconds

Initial velocity of car = 36 km/hr due north
Final velocity of car = 36 km/hr due west

The magnitude of change in velocity:

(since velocity is a vector, so direction has to be taken into account)

Acceleration = (Change in Velocity) / Time 

= [36(√2) * 5/18] / 5
= 2(√2) m/s² in South West

Time taken by first drop to cover 5 cm (u = 0), is: 
h = 1/2 gt²
5 = 1/2 x 10 x t²
t = 1 sec
Hence, the interval is 0.5 sec for each drop.
Now distance fallen by second drop in 0.5 sec
h₁ = 1/2 gt²
= 1/2 x 10 x (0.5)²         
= 5 x 0.25 
= 1.25 m
Height above the ground ( of 2^nd drop) = 5 - 1.25 = 3.75 m

The motion of the particles is roughly sketched in figure. By symmetry they will meet at the centroid O of the triangle. As the speed of all the particles is equal they will cover equal distance in any given interval of time.
If we join the instantaneous position of the particle at any time, the particles will form an equilateral triangle of same centroid as initial triangle.

Let us consider the motion of any one particle say A. At any instant, its velocity makes an angle 30^∘ with line AO.

The component of the velocity along AO is vcos30^∘. This component will be equal to the rate of change of distance between A and O.
At t=0, distance between A and O

At time t=T, the separation between A and O is zero. Hence, time taken for AO to become zero,

ac = v²/r→ constant in magnitude if v is constant.

Extension in the spring is


From Fig.

Now component of velocity along the plane becomes zero at point B

At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then