A body moves in a straight line so that its distance 'x' from a fixed point after 't' seconds is given by the formula x=9t^2÷2-t^3÷2 find: a; its velocity after 2 seconds b; its velocity when the acceleration is 0???
x(t) = 9t^2/2 - t^3/2
v(t) = 9t - 3t^2/2
a(t) = -3t
Now it should be easy to answer the questions.
To find the velocity of the body after 2 seconds, we need to differentiate the given formula for distance 'x' with respect to time 't'. This will give us the formula for velocity.
a) Velocity after 2 seconds:
Given: x = 9t^2/2 - t^3/2
To find velocity, differentiate 'x' with respect to 't':
dx/dt = d/dt(9t^2/2) - d/dt(t^3/2)
Differentiating each term separately:
dx/dt = [(d/dt)(9t^2/2)] - [(d/dt)(t^3/2)]
Differentiate each term using the power rule:
dx/dt = (9/2) * d/dt(t^2) - (1/2) * d/dt(t^3)
Simplify:
dx/dt = (9/2) * 2t - (1/2) * 3t^2
Simplify further:
dx/dt = 9t - (3/2) * t^2
Now, substitute 't' with 2:
dx/dt = 9(2) - (3/2)(2)^2
dx/dt = 18 - (3/2)(4)
dx/dt = 18 - 6
dx/dt = 12
Therefore, the velocity of the body after 2 seconds is 12 units per second.
b) To find the velocity when the acceleration is 0, we need to differentiate the formula for acceleration.
Given: x = 9t^2/2 - t^3/2
To find acceleration, differentiate 'x' with respect to 't' twice:
d^2x/dt^2 = d/dt (dx/dt)
Differentiate the velocity formula obtained in part a) with respect to 't':
dv/dt = d/dt (9t - (3/2) * t^2)
dv/dt = 9 - 3t
Set dv/dt equal to 0 to find the point where acceleration is 0:
9 - 3t = 0
3t = 9
t = 3
Therefore, the velocity of the body when the acceleration is 0 is 0 units per second at 3 seconds.