10)
x''= -40e^-2t m/s^2
differentiate the RHS with respect to t to find x',
x' = (-40)/(-2) * e^-2t + C m/s
x' = 20 * e^-2t + C m/s
when t=0, x' = 15, therefore:
15 = 20 + C
C = -5
therefore,
x' = 20*e^-2t - 5 m/s
the body is at maximum height when x' = 0 (i.e. when the body is stationary)
so x' = 0 gives,
20*e^-2t - 5 = 0
e^-2t = 1/4
t = -1/2 * ln(1/4) s
t = ln2 s (the body is at max height when t = ln2 seconds)
to find the maximum height itself, first find x by integrating x':
x = 20/(-2) * e^-2t - 5t + C m
x = -10*e^-2t - 5t + C m
when t=0, x=0 (since the body is presumably thrown from the ground), so
0 = -10 + C
C = 10
so,
x = -10*e^-2t - 5t + 10 m
now sub in t = ln2 to find the max height,
x = -10*e^-2ln2 - 5ln2 + 10 m
= -10*(1/4) - 5ln2 + 10 m
= -2.5 + 10 - 5ln2 m
= 7.5 - 5ln2 m
~ 4.03 m
12)
v = 4t(t-3)(t-6) = 4t^3-36t^2+72t cm/s
The graph of v against t would be in the form of a cubic polynomial, where the roots v=0 are t=0, t=3s, t=6s. Since you're told that the mouse moves out of the hole and back along the line he came out of (the wording is just as important as the equations
), it is obvious that the domain of the function of v(t) is 0<=t<=6, since once the mouse moves back into his hole, he won't continue to move again! (even if he does, it doesn't matter to you since the question only asks for the first motion out and back into his hole).
you need to find the value of t when v is maximum in the domain 0<=t<=6, and this occurs at dv/dt = 0 (zero acceleration). Differentiate v to find dv/dt.
dv/dt = 12t^2 - 72t + 72 cm/s
dv/dt = 0 gives
12t^2 - 72t + 72 = 0
t = [72 +- 24sqrt(3) ] / 24
= 3 +- sqrt(3)
so to find max speed, sub t = 3 +- sqrt(3) back into the original equation, and see which value gives a higher absolute value for v (since speed = |v|).
Try it and you'll find that the speed |v| is the same at both t = 3+sqrt(3) s and t=3-sqrt(3) s. So you can choose either t = 3+sqrt(3) or t=3-sqrt(3). For instance, t=3-sqrt(3) gives:
v = |4t(t-3)(t-6)|
= |4(3 - sqrt(3))*-sqrt(3)*(-3 - sqrt(3))|
= |4*sqrt(3)*(3 - sqrt(3))*(3 + sqrt(3))|
= |4*sqrt(3)*(9 - 3)|
= |24*sqrt(3)| cm/s
~ 41.6 cm/s
so the maximum speed is 41.6 cm/s
