x = 6t - t^3...?

Question

A house undergoes a displacement x (in mm) given by the formula

x = 6t - t^3

following an earthquake, where t is the time (in seconds) after the shock.

i) What is dx/dt?
ii) What is the time t when the displacement is at maximum?
iii) What is the maximum displacement?

Answer 1

dx/dt = 6 - 3t^2

6 - 3t^2 = 0

6 = 3t^2

2 = t^2

+/- SQRT(2) = t

The time is approximately 1.414 seconds

Use the positive answer (throw out the negative)

x = 6 SQRT (2) - ( SQRT(2))^3

The maximum displacement is approximately 5.657 mm

Answer 2

x = 6t - t^3

i) dx/dt = 6 - 3t^2

ii) Displacement is maximum when dx/dt = 0
So 6 - 3t^2 = 0
6 = 3t^2
2 = t^2
t = 1.41s

Check for maximum:
d2x/dt2 = -6t < 0 => maximum

iii) Maximum displacement
= 6t - t^3
= 6(1.41) - (1.41)^3
= 5.66mm

Answer 3

x = 6t - t^3.
dx/dt = 6 - 3t^2

ii) To find maximum point, set dx/dt to 0 to get:
6 - 3t^2 = 0
= t^2 = 2
Therefore, t = + or minus sqrt(2).

Now to find the maximum point, put the two values of t in d2x.dt^2.

d^2x/dt^2 = - 6t
When t = root 2, the second derivative is negative, therefore, it is maximum. So the maximum value is when t = root 2

iii) Now substitute the value of t in the original function to get:

x = 6 * root 2 - (root 2)^3
= 5.657 mm

Hope this helps.

Answer 4

(i)
dx/dt = 6 - 3t^2

(ii)
Displacement is maximum when dx/dt = 0
6 - 3t^2 = 0
3t^2 = 6
t^2 = 2
t = sqrt(2) (negative time not possible)
d^2x/dt^2 = -6t < 0, proving the displacement is a maximum and not a minimum.

(ii)
Maximum displacement is:
6sqrt(2) - 2sqrt(2) = 4sqrt(2).

Answer 5

i)
dx/dt = 6 - 3.t²

ii)
6 - 3t² = 0
t² = 2
t = √2 sec (at turning point)

ii)
max. displacement :-
x = 6.√2 - (√2)³
x = 6.√2 - 2√2
x = 4.√2 mm

Answer 6

dx/dt= 6 - 3t^2
x" = -6t
put dx/dt = 0
t = sqrt(2)
Since x'' is negative, so at t = sqrt(2), x is maximum
So the maximum value of x = 6 * 2^1/2 - 2^3/2= 4 sqrt(2)

Answer 7

dx/dt=6-3t^2 mm/s
dx/dt=0
6-3t^2=0 t=sqrt2 seg The sign of x´is 0++++++sqrt2 ---------
so at x= sqrt2 we have a local maximum which is also the place of the absolute maximum
x(sqrt2)=6sqrt 2-2sqrt2=4sqrt2 mm ( maximum displacement)

Answer 8

dx/dt = 6 - 3t^2
t is max when dx/dt = 0 => 3t^2 = 6 => t=+-2^1/2
x(2^1/2) < x(-2^1/2) => at t=-2^1/2
The max displacement is: x(-2^1/2)

Answer 9

dx/dt = 6 - 3t^2

0 = 6 - 3t^2
6 = 3t^2
2 = t^2
t = 1.414

x = 6(1.414) - 1.414^3
x = 8.484 - 2.828
x = 5.656

Answer 10

x=6t-t^3
dx/dt=3t^2+6