An object is moving along the x direction with a velocity (in mks units) given by
vx = 3 t^4 − 12 t^3 + 12 t^2 + v0x
Answer the following.
(a) At what three times is the object moving with a constant velocity?
(b) Find the velocity of the object in terms of v0x at each of the above three times.
(c) If the velocity at the middle time is 2 times that of the velocity at the longest time,
how fast was the object moving at t = 0?
(d) How far did the object move in the first second of its motion? State the exact value
here, not a decimal approximation
To answer these questions, we need to find the times at which the object is moving with a constant velocity, the velocity at those times in terms of v0x, and the initial velocity at t = 0. We also need to determine the distance covered in the first second of motion.
(a) To find the times when the object is moving with a constant velocity, we can differentiate the given velocity function with respect to time and set it equal to zero. This will give us the times when the object's acceleration is zero, indicating constant velocity. Let's find the derivative of vx:
vx = 3t^4 - 12t^3 + 12t^2 + v0x
Differentiating both sides with respect to time:
dvx/dt = 12t^3 - 36t^2 + 24t
Setting dvx/dt equal to zero:
12t^3 - 36t^2 + 24t = 0
Factoring out common terms:
12t(t^2 - 3t + 2) = 0
Simplifying:
t(t - 1)(t - 2) = 0
So the times at which the object is moving with constant velocity are t = 0, t = 1, and t = 2.
(b) To find the velocity of the object at each of the above three times, we substitute these values back into the velocity function vx:
At t = 0:
vx = 3(0)^4 - 12(0)^3 + 12(0)^2 + v0x
vx = v0x
At t = 1:
vx = 3(1)^4 - 12(1)^3 + 12(1)^2 + v0x
vx = 3 - 12 + 12 + v0x
vx = 3 + v0x
At t = 2:
vx = 3(2)^4 - 12(2)^3 + 12(2)^2 + v0x
vx = 48 - 96 + 48 + v0x
vx = 0 + v0x
vx = v0x
So at t = 0 and t = 2, the velocity of the object is v0x, and at t = 1, the velocity is 3 + v0x.
(c) Given that the velocity at the middle time (t = 1) is 2 times that of the velocity at the longest time (t = 2), we can write the equation:
3 + v0x = 2(v0x)
Simplifying:
3 + v0x = 2v0x
3 = v0x
Therefore, the object was moving at a velocity of 3 m/s at t = 0.
(d) To find the distance covered in the first second of motion, we need to calculate the displacement. Displacement equals the integral of velocity over the interval t = 0 to t = 1. Let's calculate it using the velocity function:
Displacement = ∫(vx) dt (from 0 to 1)
Integrating:
Displacement = ∫(3t^4 - 12t^3 + 12t^2 + v0x) dt (from 0 to 1)
Displacement = t^5 - 3t^4 + 4t^3 + v0x t | (from 0 to 1)
Displacement = 1 - 3 + 4 + v0x - 0
Displacement = 2 + v0x
So, the object moved a distance of 2 + v0x meters in the first second of its motion.