6) A particle P is moving along the x-axis with velocity v=4t-2t^2. When t=0, P is at x=3. Find a) the position of P when t=2 b)the maximum velocity attainted by P c) the distance OP when the velocity is maximum
7)A particle P os moving along the x-axis. When t=0, the velocity of P is 4.5 m/s. At time ts the acceleration of P is given by (3t-6) m/s^2. Find the time when the particle returns to its starting point.
(8) A particle moves along the x-axis with velocity at time t given by v=5t^2+2t. Find (a) the distance it moves in the 2nd second (b) the distance it moves in the 4th second.
I did some of question 6 but got confused. Thanks.
x Turn on thread page Beta
Kinematics Help! watch
- Thread Starter
- 13-03-2013 15:09
- 14-03-2013 23:18
For part a) integrate v with respect to t and to get x in terms of t (use boundary conditions t=0, x=3 to find the constant of integration)
Part b) when the velocity is max, the acceleration is 0 (if velocity is max, then its not changing), so differentiate the velocity to get acceleration, then put acceleration equal to 0 to find the value of t that v=vmax; then sub that t value into v to find the velocity at that time.
c) sub your value for t (when v=vmax) into the displacement equation to get the displacement at that time.
For 7: you have acceleration, and the question asks for the time when its at the start point.... So you need an equation relating x to t....integrate acceleration twice to find displacement (remember to use the boundary conditions to get the constants of integration)
For 8: integrate v to find x, then sub in the numbers the question asks you for