A particle moves along a straight line with a velocity, v m s-1, given by v =4-8sin2t, where t is the time in seconds after passing a fixed point O. Determine the range of values of v
To determine the range of values of velocity, we need to find the minimum and maximum values of v based on the given equation v = 4 - 8sin^2(t).
The maximum value of sin^2(t) is 1, and the minimum value is 0. Therefore, the maximum value of v occurs when sin^2(t) is at its maximum, which is 1. Plugging this into the equation, we get:
v = 4 - 8(1)
v = 4 - 8
v = -4
Similarly, the minimum value of v occurs when sin^2(t) is at its minimum, which is 0. Plugging this into the equation, we get:
v = 4 - 8(0)
v = 4 - 0
v = 4
So, the range of values of velocity v is -4 m/s to 4 m/s.
To determine the range of values of v, we need to find the minimum and maximum values of v.
Given that v = 4 - 8sin^2(t), we can see that the term 8sin^2(t) will vary between 0 and 8, since the sine function oscillates between -1 and 1.
Therefore, the minimum value of v occurs when sin^2(t) is maximum, which is 1. So, when sin^2(t) = 1, the minimum value of v is:
v_min = 4 - 8 * 1
= 4 - 8
= -4
Similarly, the maximum value of v occurs when sin^2(t) is minimum, which is 0. So, when sin^2(t) = 0, the maximum value of v is:
v_max = 4 - 8 * 0
= 4 - 0
= 4
Hence, the range of values of v is from -4 to 4 (inclusive).
Since the max of sin(anything) is +1 and the minimum is -1
v has a max of 4 -(-8) or 12 and a min of 4 - 8(1) or -4, so
-4 ≤ v ≤ 12