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# Velocity = 3cos(t)-2sin(t)?

t stands for time:

V(t) = 3cos(t)-2sin(t)

When is the speed the fastest?

### 2 Answers

- 1 decade agoFavorite Answer
Express V(t) in the form R cos (t+a)

3 cos t - 2 sin t = R cos (t+a) = R cos t cos a - R sin t sin a

3 cos t = R cos t cos a

3 = R cos a

2 sin t = R sin t sin a

2 = R sin a

R^2 (sin^2 a + cos^2 a) = 2^2 +3^2

R = sqrt (13)

sin a = 2 / sqrt(13) cos a = 3 / sqrt(13)

a=0.588 radians

Velocity = 3 cos (t) - 2 sin (t) = SQRT(13) * cos (t+0.588)

Cos (x) has a maximum value of 1, occurring at x=0, 2pi, 4pi, ... and minimum value of -1, occurring at x= pi, 3pi, ...

Therefore the speed will be at the maximum of +sqrt(13)

at:

2pi-0.588 = 5.70

4pi-0.588 = 11.98

6pi-0.588

and so on for even multiples of pi.

The speed will be at the minimum of -sqrt(13) at:

3pi-0.588=8.84

5pi-0.588=15.12

7pi-0.588

and so on for odd multiples of pi.

- sahsjingLv 71 decade ago
V(t)

= 3cos(t)-2sin(t)

= √13[(3/√13)cos(t) - (2/√13)sin(t)]

= -√13sin[t-arctan(3/2)]

|v| (max) = √13

Solve |v(t)| = √13,

t = pi/2 + arctan(3/2) + n*pi = 2.5536 + n*pi, n can be any integer.

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Reason:

|v| (max) = the amplitude of v(t).

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r_k_stan,

Your answer is the same as mine.