A particle is moving in a straight line. It's displacement at any inst...
Answer:
Given:
- Displacement of the particle at any instant is x=10t 15t^3
- Time interval t=0 to t=2 s
Part 1: Average acceleration
- The formula for average acceleration is a_avg= (v_f - v_i)/t where v_f is the final velocity, v_i is the initial velocity and t is the time interval.
- To find the average acceleration for the given time interval, we need to find the velocity at the beginning and end of the time interval.
- The velocity of the particle at any instant is given by v=dx/dt where x is the displacement.
- Differentiating x=10t + 15t^3 with respect to t, we get v=10+45t^2
- The initial velocity at t=0 is v_i=10 m/s
- The final velocity at t=2s is v_f=10+45(2)^2=190 m/s
- Substituting these values in the formula for average acceleration, we get a_avg= (190-10)/2=90 m/s^2
Part 2: Instantaneous acceleration
- The instantaneous acceleration at any instant is given by a= dv/dt where v is the velocity.
- Differentiating v=10+45t^2 with respect to t, we get a=90t
- Substituting t=2s, we get a=180 m/s^2
Explanation:
- The given displacement equation x=10t+15t^3 represents a particle moving in a straight line.
- To find the average acceleration, we used the formula a_avg= (v_f - v_i)/t where v_f is the final velocity, v_i is the initial velocity and t is the time interval.
- To find the instantaneous acceleration, we used the formula a= dv/dt where v is the velocity.
- We found the velocity by differentiating the displacement equation with respect to time.
- By substituting the values in the formulas, we found the average acceleration to be 90 m/s^2 and the instantaneous acceleration at t=2s to be 180 m/s^2.
A particle is moving in a straight line. It's displacement at any inst...
1:- dx/dt=10t+15t^3
v=10+30t^2
dv/dt=10+30t^2
a=60t
t=2
a=120m/s^2
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