A particle moves along a straight line with a velocity, v ms/s, given by v = 4 - 8 sin 2t where t is the time in seconds after passing a fixed point O.
1. Determine the range of values of v.
well, sin(2t) varies from -1 to 1.
To determine the range of values of v, we need to find the maximum and minimum values of the function v = 4 - 8 sin 2t.
The maximum and minimum values of the sine function are 1 and -1, respectively. So, the maximum and minimum values of 8 sin 2t are 8 and -8, respectively.
The function v = 4 - 8 sin 2t is the sum of a constant value (4) and the function 8 sin 2t. Therefore, the maximum and minimum values of v are:
Maximum value: 4 + 8 = 12 m/s
Minimum value: 4 - 8 = -4 m/s
Hence, the range of values of v is -4 m/s to 12 m/s.
To determine the range of values of v, we need to find the maximum and minimum values that v can reach.
First, let's analyze the given equation for v:
v = 4 - 8 sin (2t)
The sine function sin(2t) oscillates between -1 and 1. Therefore, the maximum and minimum values of v occur when sin(2t) is equal to 1 and -1, respectively.
When sin(2t) = 1, v = 4 - 8(1) = 4 - 8 = -4
When sin(2t) = -1, v = 4 - 8(-1) = 4 + 8 = 12
Therefore, the range of values for v is -4 ≤ v ≤ 12.