The position of a toy locomotive mobing on a straight track along the x-axis is given the equation x=t^3-6t^2 +9t where x is in metters and t is in seconds. The net force on the locomotive is equal to zero when t is eqaul to
a) zero
b)2 s
c)3 s
d)4s
e)5s
Would the answer be b)2 s? I found the double derivative of the equation and set x to 0 and solved for t.
f'= 3t^2 - 12t + 9
f"=6t - 12
yes, t=2
Yes, you are correct. The answer is b) 2 s.
You found the double derivative (acceleration) of the position function and set it to zero, which is the condition for the net force to be zero (F = ma, and F = 0 when a = 0). From there, you solved for t and found t=2, which corresponds to option b.
Yes, your answer is correct. To find the time when the net force on the locomotive is zero, you need to find the values of t when the second derivative of the position function equals zero.
As you correctly calculated, the second derivative f"= 6t - 12. Setting it equal to zero:
6t - 12 = 0
6t = 12
t = 2
Therefore, the net force on the locomotive is equal to zero when t is equal to 2 seconds. Thus, the correct answer is b) 2s.
Yes, you are correct. To find the time when the net force on the locomotive is equal to zero, we need to find the values of t that make the second derivative of the position equation, f'', equal to zero.
As you correctly found, the second derivative is f''(t) = 6t - 12. Setting this equal to zero and solving for t:
6t - 12 = 0
6t = 12
t = 2
So, when t is equal to 2 seconds, the net force on the toy locomotive is equal to zero. Therefore, the answer is option b) 2s.