A particle starts from rest and moves in a straight line such that its acceleration after t secs is given by \(a = (3t - 2) ms^{-2}\). Find the distance cov...
A particle starts from rest and moves in a straight line such that its acceleration after t secs is given by \(a = (3t - 2) ms^{-2}\). Find the distance covered after 3 secs.
Answer Details
To solve this problem, we need to use the kinematic equation:
$$s = ut + \frac{1}{2}at^2$$
where:
- s is the distance covered
- u is the initial velocity (which is zero in this case)
- a is the acceleration
- t is the time taken
We are given that the acceleration is given by \(a = (3t-2)ms^{-2}\). Integrating this with respect to time, we get:
$$v = \int a dt = \int (3t-2) dt = \frac{3}{2}t^2 - 2t + C$$
where v is the velocity of the particle at time t, and C is a constant of integration. Since the particle starts from rest, we have v = 0 when t = 0, so:
$$0 = \frac{3}{2}(0)^2 - 2(0) + C$$
which gives C = 0. Therefore, the velocity of the particle at time t is given by:
$$v = \frac{3}{2}t^2 - 2t$$
Now, we can use the equation for distance covered to find the distance covered by the particle after 3 seconds:
$$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3t-2)t^2 = \frac{3}{2}t^3 - t^2$$
So, when t = 3, we have:
$$s = \frac{3}{2}(3)^3 - (3)^2 = \frac{27}{2} - 9 = \frac{9}{2}m$$
Therefore, the distance covered by the particle after 3 seconds is \(\frac{9}{2}m\). Thus, the correct option is:
- \(\frac{9}{2}m\)