Question
a particle moves along a line so that at time t where 0<=t<=pi, its position is given by s(t)=-4sint-t/2+10. What is the acceleration of the particle the first time its velocity is zero?
Answers
Bot
To find the velocity of the particle, we need to take the derivative of the position function:
v(t) = s'(t) = -4cos(t) - 1/2
To find the time at which the velocity is zero, we need to solve the equation v(t) = 0:
-4cos(t) - 1/2 = 0
cos(t) = -1/8
Since 0 <= t <= pi, we know that t must be in the second or third quadrant, where cosine is negative. We can use an inverse cosine function to find the solution in the second quadrant:
t = 2*pi - arccos(-1/8) = 5.74 (rounded to two decimal places)
Now that we know the time at which the velocity is zero, we can find the acceleration by taking the derivative of the velocity function:
a(t) = v'(t) = 4sin(t)
At t = 5.74, sin(t) is approximately -0.888, so the acceleration is:
a(5.74) = 4(-0.888) = -3.55 (rounded to two decimal places)
Therefore, the acceleration of the particle the first time its velocity is zero is approximately -3.55.
v(t) = s'(t) = -4cos(t) - 1/2
To find the time at which the velocity is zero, we need to solve the equation v(t) = 0:
-4cos(t) - 1/2 = 0
cos(t) = -1/8
Since 0 <= t <= pi, we know that t must be in the second or third quadrant, where cosine is negative. We can use an inverse cosine function to find the solution in the second quadrant:
t = 2*pi - arccos(-1/8) = 5.74 (rounded to two decimal places)
Now that we know the time at which the velocity is zero, we can find the acceleration by taking the derivative of the velocity function:
a(t) = v'(t) = 4sin(t)
At t = 5.74, sin(t) is approximately -0.888, so the acceleration is:
a(5.74) = 4(-0.888) = -3.55 (rounded to two decimal places)
Therefore, the acceleration of the particle the first time its velocity is zero is approximately -3.55.
oobleck
AAAaannndd the bot gets it wrong yet again!
The first time v(t) = 0 is at t=1.696
If cos(t) = -1/8, then sin(t) = -√63/8
so 4 sin(t) = -√63/2 = -3.96
The first time v(t) = 0 is at t=1.696
If cos(t) = -1/8, then sin(t) = -√63/8
so 4 sin(t) = -√63/2 = -3.96
Bot
I apologize for my previous incorrect response. You are correct that the first time the velocity is zero is at t = 1.696, and that the acceleration at this time is -3.96 (rounded to two decimal places). Thank you for bringing this to my attention and I will make sure to double-check my calculations in the future.