Home
Class 11
PHYSICS
A: An object moving with a velocity of m...

A: An object moving with a velocity of magnitude 10 m/s is subjected to a uniform acceleration `2 m//s^(2)` at right angle to the initial motion. Its velocity after 5 s has a magnitude nearly 14 m/s
`rarr rarr rarr`
R: The equation `vec(v) = vec(u) + vec(a)t` can be applied to obtain `vec(v)` if `vec(a)` is constant

A

If both A and B are true and R is the correct explanation of A

B

If both A and R are true but R is not correct explantion of A

C

If A is true but R is false

D

If A is false but R is true

Text Solution

Verified by Experts

The correct Answer is:
A
Promotional Banner

Topper's Solved these Questions

  • MOTION IN A STRAIGHT LINE

    GRB PUBLICATION|Exercise Matrix -match|4 Videos

  • MOTION IN A STRAIGHT LINE

    GRB PUBLICATION|Exercise Integer|5 Videos

  • MOTION IN A STRAIGHT LINE

    GRB PUBLICATION|Exercise More than one choice is correct|16 Videos

  • FRICTION AND CIRCULAR MOTION

    GRB PUBLICATION|Exercise Comprehension type|11 Videos

  • MOTION IN TWO AND THREE DIMENSIONS

    GRB PUBLICATION|Exercise All Questions|210 Videos

Similar Questions

Explore conceptually related problems

A particle moving with a velocity equal to 0.4 m//s is subjected to an acceleration of 0.15 m//s^(2) for 2 s . in a direction at the right angle to its direction of motion. The resultant velocity is

A body is moving with a velocity of 15 m//s . If the motion is uniform , what will be the velocity after 10 s ?

A particle is moving with a velocity of vec(v)=(3hat(i)+4that(j))m//s . Find the ratio of rangential acceleration to that of total acceleration at t=1sec

Assertion : A body of mass 4 kg has an initial velocity 5hat(i)ms^(-1) . It is subjected to a force of 4hat(j)N . The displacement of body from origin after 4 s will be 21.5 m . Reason : The equation v=u+a t can be applied to obtain v if a is constant.

A particle starts moving with velocity "vecV=hat i(m/s)", and acceleration "vec a=(hat i+hat j)(m/s^(2))".Find the radius of curvature of the particle at "t=1sec".

Velocity of a particle at time t=0 is 2 m//s. A constant acceleration of 2 m/s^2 acts on the particle for 2 s at an angle of 60^@ with its initial velocity. Find the magnitude of velocity and displacement of particle at the end of t=2s.

An object has velocity vec v = (i+2hatj+3hatk)m//s . A constant acceleration of the object for which its speed starts decreasing is:

A particle has initial velocity (2 vec(i)+3 vec(j)) and acceleration (0.3 vec (i)+0.2 vec (j)) . The magnitude of velocity after 10 seconds will be

The velocity of an object at t =0 is vecv_(0) =- 4 hatj m/s . It moves in plane with constant acceleration veca = ( 3hati + 8 hatj) m//s^(2) . What is its velocity after 1 s?

if a=3t^(2)+2t , initial velocity is 5 m//s . Find the velocity at t=4s. The motion is in straight line, a is acceleration in m//s^(2) and t is time in seconds.