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phospho
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Just two questions, if anyone has the time to check over my answers (no mark scheme is available).
Q1. A particle P moves on the x-axis. The acceleration of P at time t seconds, t ≥ 0 is (3t+5) ms^-2 in the positive x-direction. When t = 0, the velocity of P is 2 ms^-1 in the positive x-direction. When t = T, the velocity of P is 6ms^-1 in the positive x direction. Find the value of T.
Working:
integrating the acceleration to find the velocity, I get ((3t^2)/2) + 5t + C, finding the constant by using v = 2 when t = 0, the constant is 0
so v = 6, when t = T
6 = (3T^2)/2 + 5T + 2
solving this to get T = 3/2 or T = -4
as t ≥ 0, T = 3/2
Q9.
A particle P moves along the x-axis. At time t seconds the velocity of P is v ms^-1 in the positive x-direction, where v = 3t^2 - 4t + 3. When t = 0, P is at the origin 0. Find the distance of P from O when P is moving with minimum velocity.
Taking the derivative of v, we get 6t - 4, letting that = 0 we get t = 2/3.
then we integrate the velocity to get an equation of the displacement, and P is at O when t = 0 the constant is 0, so the displacement: x = t^3 - 2t^2 + 3t
substuting t = (2/3) we get the distance to be (38/27)
If anyone could actually check over these (I need correct answers!) it'd be really good help!
Thank yoU!
Q1. A particle P moves on the x-axis. The acceleration of P at time t seconds, t ≥ 0 is (3t+5) ms^-2 in the positive x-direction. When t = 0, the velocity of P is 2 ms^-1 in the positive x-direction. When t = T, the velocity of P is 6ms^-1 in the positive x direction. Find the value of T.
Working:
integrating the acceleration to find the velocity, I get ((3t^2)/2) + 5t + C, finding the constant by using v = 2 when t = 0, the constant is 0
so v = 6, when t = T
6 = (3T^2)/2 + 5T + 2
solving this to get T = 3/2 or T = -4
as t ≥ 0, T = 3/2
Q9.
A particle P moves along the x-axis. At time t seconds the velocity of P is v ms^-1 in the positive x-direction, where v = 3t^2 - 4t + 3. When t = 0, P is at the origin 0. Find the distance of P from O when P is moving with minimum velocity.
Taking the derivative of v, we get 6t - 4, letting that = 0 we get t = 2/3.
then we integrate the velocity to get an equation of the displacement, and P is at O when t = 0 the constant is 0, so the displacement: x = t^3 - 2t^2 + 3t
substuting t = (2/3) we get the distance to be (38/27)
If anyone could actually check over these (I need correct answers!) it'd be really good help!
Thank yoU!