maximum velocity

Question

At a carnival, an animated target oscillates in a straight line from a fixed point, P, such that after t seconds, its position, in meters per second, is given by S(t)= cos(t) - cos (2t). Determine the maximum velocity that the target attains

Answer 1

velocity is the derivative of position:

v(t) = -sin(t) + 2sin(2t)

max velocity occurs when acceleration is 0. one more derivative

a(t) = -cos(t) + 4cos(2t)

a(t) = -cos(t) + 4(2cos^2(t) - 1)

for a = 0:

8cos^2(t) - cos(t) - 4 = 0

solve the quadratic equation

t = arccos((1+sqrt(129))/16)

this is the INSTANT the max vecloity is attained, not the velocity itself

so substitute back in

vmax = 1.327181022458534431767601400963

Answer 2

If position is s(t) then velocity is s'(t) or ds/dt (the derivative of position).

For s(t) = cos(t) - cos(2t) then s'(t) = v(t) = -sin(t) + 2sin(2t)

To find the maximum of any function, we take it's derivative, set it equal to zero and solve.

So, for v(t) = -sin(t) + 2sin(2t), v'(t) = -cos(t) + 4cos(2t)

Setting it equal to zero:

-cos(t) + 4cos(2t) = 0

Using the double angle identity cos(2x) = 2cos²(x) – 1, our equation becomes:

-cos(t) + 4(2cos²(t) – 1) = 0

8cos²(t) - cos(t) - 4 = 0

Let cos(t) = w, then the equation becomes 8w² - w - 4 = 0

Using the quadratic formula, w = [1 ± √(1² -4(8)(-4))]/(2•8) = (1 ± √129)/16 = 0.77236 and -0.64736

Back substituting, we get cos(t) = 0.77236 and cos(t) = -0.64736

So t = cos^-1(0.77236) and t = cos^-1(-0.64736)

To find the value of "t", we must consider the unit circle. Cosine will be positive in Q1 and in Q4 and negative in Q2 and Q3.

The Q1 value we can get from a calculator as cos^-1(0.77236) which gives t = 0.688 seconds

The Q4 value, we can get from a calculator using 2π - cos^-1( 0.77236) or t = 5.595 seconds

The Q2 value, we can get from a calculator using cos^-1( -0.64736) or t = 2.275 seconds

And the Q3 value, we can get from a calculator using π + cos^-1( 0.64736) or t = 4.008 seconds

Since sin(t) has a period of 2π and sin(2t) has a period of π, the values of t will repeat after our Q4 value so these are all the minimums and maximums.

Since we need to determine the maximum velocity, we might has well just plug in these values into v(t) to see what they are instead of doing a derivative test to check for max/min values.

So for t = 0.688 s, t = 2.275 s, t = 4.008 s, and t = 5.595 s we get :

v(0.688) = -sin(0.688) + 2sin(2•0.688) = 1.327 m/s

v(2.275) = -sin(2.275) + 2sin(2•2.275) = -2.736 m/s

v(4.008) = -sin(4.008) + 2sin(2•4.008) = 2.736 m/s

v(5.595) = -sin(5.595) + 2sin(2•5.595) = -1.327 m/s

So the maximum velocity occurs at t = 4.008 seconds and has a value of 2.736 m/s