A particle P moves in a straight line such that t seconds after passing a fixed point Q, its velocity is given by the equation V=2t^2-10t+12 , find
(a) The values of t when P is instantaneously at rest
(b) An expression for the distance moved by P after t seconds
(c) The total distance traveled by P in the first 3 seconds after passing point Q
(d). The distance of P from Q when acceleration is zero
(a) To find when P is instantaneously at rest, we need to find the values of t when the velocity V is equal to zero. So we set 2t^2 - 10t + 12 = 0 and solve for t.
Using the quadratic formula, t = (10 ± √(100 - 96)) / 4 = (5/2) ± (1/2)
So, t = 2 or t = 3.
Therefore, P is instantaneously at rest after 2 seconds and after 3 seconds.
(b) To find an expression for the distance moved by P after t seconds, we integrate the velocity function V with respect to time t.
d = ∫V dt = ∫(2t^2 - 10t + 12) dt = (2/3)t^3 - 5t^2 + 12t + C
where C is an arbitrary constant of integration.
To find the value of C, we use the fact that P passes the point Q at time t = 0, so d = 0 when t = 0.
Therefore, 0 = (2/3)(0)^3 - 5(0)^2 + 12(0) + C
C = 0.
So, the expression for the distance moved by P after t seconds is:
d = (2/3)t^3 - 5t^2 + 12t
(c) The total distance traveled by P in the first 3 seconds after passing point Q is:
d(0 to 3) = (2/3)(3)^3 - 5(3)^2 + 12(3) = 9
Therefore, P travels a distance of 9 units in the first 3 seconds after passing point Q.
(d) To find the distance of P from Q when acceleration is zero, we need to find the time t when the acceleration is zero. The acceleration is the derivative of the velocity function V:
a = dV/dt = 4t - 10
Setting a = 0, we get:
4t - 10 = 0
t = 2.5
So, the acceleration is zero at time t = 2.5 seconds.
To find the distance at time t = 2.5 seconds, we plug t = 2.5 into the expression for the distance moved by P:
d = (2/3)(2.5)^3 - 5(2.5)^2 + 12(2.5) = 6.875
Therefore, the distance of P from Q when acceleration is zero is 6.875 units.
(a) A particle is instantaneously at rest when its velocity is equal to 0. Thus, we need to find the values of t when V=0:
0 = 2t^2 - 10t + 12
This is a quadratic equation which we can solve using the quadratic formula or factoring. Factoring gives:
0 = (2t - 6)(t - 2)
This gives us two possible values for t:
2t - 6 = 0 => t = 3
t - 2 = 0 => t = 2
Thus, particle P is instantaneously at rest at t = 2 seconds and t = 3 seconds.
(b) To find the distance moved by P after t seconds, we need to find the displacement of the particle, which can be found by integrating the velocity function:
s(t) = ∫(2t^2 - 10t + 12) dt
s(t) = (2/3)t^3 - 5t^2 + 12t + C
To find the constant C, we know that at t=0, the particle is at point Q, which we can assign a displacement of 0:
s(0) = 0 = (2/3)(0)^3 - 5(0)^2 + 12(0) + C => C = 0
Thus, the expression for the distance moved by P after t seconds is:
s(t) = (2/3)t^3 - 5t^2 + 12t
(c) To find the total distance traveled by P in the first 3 seconds, we need to evaluate the displacement expression at t = 3:
s(3) = (2/3)(3)^3 - 5(3)^2 + 12(3)
s(3) = (2/3)(27) - 5(9) + 36
s(3) = 18 - 45 + 36
s(3) = 9
Thus, the total distance traveled by P in the first 3 seconds is 9 units.
(d) The distance of P from Q when acceleration is zero can be found by first finding the expression for acceleration by differentiating the velocity function:
a(t) = dV/dt = 4t - 10
Now, we set a(t) equal to 0 to find when acceleration is zero:
0 = 4t - 10
t = 10/4 = 5/2 = 2.5
Now, we use the displacement expression to find the distance of P from Q at t = 2.5:
s(2.5) = (2/3)(2.5)^3 - 5(2.5)^2 + 12(2.5)
s(2.5) = (2/3)(15.625) - 5(6.25) + 30
s(2.5) = 10.417 - 31.25 + 30
s(2.5) = 9.167
Thus, the distance of P from Q when the acceleration is zero is 9.167 units.