Home
Class 12
MATHS
A particle is thrown with 12 m/s in posi...

A particle is thrown with 12 m/s in positive X direction while acceleration is `2 m//s^2` in negative X direction then distance travelled by particle between t=5 to t=7 sec is

Text Solution

Verified by Experts

The correct Answer is:
2.25
Promotional Banner

Topper's Solved these Questions

  • TEST PAPER

    ALLEN |Exercise CHEMISTRY SECTION-I|20 Videos
  • TEST PAPER

    ALLEN |Exercise CHEMISTRY SECTION-II|16 Videos
  • TEST PAPER

    ALLEN |Exercise PHYSICS SECTION-I|20 Videos
  • SEQUENCE AND PROGRESSION

    ALLEN |Exercise Exercise (JA)|10 Videos
  • TEST PAPERS

    ALLEN |Exercise MATHEMATICS|136 Videos

Similar Questions

Explore conceptually related problems

A particle starts with an initial velocity 2.5 m//s along the posiive x-direction and it accelerates uniformly at the rate 0.50 m//s^2. Find the distance travelled by it in the first two seconds

A particle moves on x-axis as per equation x = (t^(3)- 9t^(2) +15t +2)m . Distance travelled by the particle between t = 0 and t = 5s is

If velocity (in ms^(-1) ) varies with time as V = 5t, find the distance travelled by the particle in time interval of t = 2s to t = 4 s.

Position of a particle moving along x-axis is given by x=2+8t-4t^(2) . The distance travelled by the particle from t=0 to t=2 is:-

The speed-time graph of a particle moving along a fixed direction is shown in figure. Obtain the distance traversed by the particle between (a) t =0 s to 10s. (b) t = 2 s to 6s. What is the average speed of the particle over the intervals in (a) and (b) ?

When a particle is undergoing motion, the diplacement of the particle has a magnitude that is equal to or smaller than the total distance travelled by the particle. In many cases the displacement of the particle may actually be zero, while the distance travelled by it is non-zero. Both these quantities, however depend on the frame of reference in which motion of the particle is being observed. Consider a particle which is projected in the earth's gravitational field, close to its surface, with a speed of 100sqrt(2) m//s , at an angle of 45^(@) with the horizontal in the eastward direction. Ignore air resistance and assume that the acceleration due to gravity is 10 m//s^(2) . There exists a frame (D) in which the distance travelled by the particle is minimum. This minimum distance is equal to :-

A particle is moving with a position vector, vec(r)=[a_(0) sin (2pi t) hat(i)+a_(0) cos (2pi t) hat(j)] . find Distance travelled by the particle in 1 sec is

A particle is travelling in a circular path of radius 4m . At a certain instant the particle is moving at 20m/s and its acceleration is at an angle of 37^(@) from the direction to the centre of the circle as seen from the particle (i) At what rate is the speed of the particle increasing? (ii) What is the magnitude of the acceleration?

Velocity of a particle varies with time as v=4t. Calculate the displacement of particle between t=2 to t=4 sec.

When a particle is undergoing motion, the diplacement of the particle has a magnitude that is equal to or smaller than the total distance travelled by the particle. In many cases the displacement of the particle may actually be zero, while the distance travelled by it is non-zero. Both these quantities, however depend on the frame of reference in which motion of the particle is being observed. Consider a particle which is projected in the earth's gravitational field, close to its surface, with a speed of 100sqrt(2) m//s , at an angle of 45^(@) with the horizontal in the eastward direction. Ignore air resistance and assume that the acceleration due to gravity is 10 m//s^(2) . Consider an observer in frame D (of the previous question), who observes a body of mass 10 kg acelerating in the upward direction at 30 m//s^(2) (w.r.t. himself). The net force acting on this body, as observed from the ground is :-