The position of a body at time t sec is s=t^3-6t^2+9t m. Find the body's acceleration each time the velocity is zero.
the velocity of a car (measured in ft/sec) "t" seconds. after starting from rest and moving in a straight line is given by f(t)=2t (0<t<20). find the displacement of a car at any time "t"
To find the acceleration when the velocity is zero, we need to determine the velocity function first. The velocity is the derivative of the position function, so let's differentiate the position function with respect to time:
s = t^3 - 6t^2 + 9t
Taking the derivative using the power rule:
v = ds/dt = 3t^2 - 12t + 9
Now, let's find the values of t where the velocity is zero by setting v = 0:
0 = 3t^2 - 12t + 9
This is a quadratic equation. We can solve it by factoring or using the quadratic formula. Let's use the quadratic formula:
t = (-(-12) ± √((-12)^2 - 4(3)(9))) / (2 * 3)
Simplifying further:
t = (12 ± √(144 - 108)) / 6
t = (12 ± √36) / 6
t = (12 ± 6) / 6
Thus, we have two possible values for t:
1) When t = (12 + 6) / 6 = 3
2) When t = (12 - 6) / 6 = 1
Now that we have the values of t when the velocity is zero, we can find the acceleration at each time. The acceleration is the derivative of the velocity function. Let's differentiate the velocity function with respect to time:
a = dv/dt = d^2s/dt^2 = d/dt(3t^2 - 12t + 9)
Taking the derivative:
a = 6t - 12
For t = 3:
a = 6(3) - 12 = 18 - 12 = 6 m/s^2
For t = 1:
a = 6(1) - 12 = 6 - 12 = -6 m/s^2
Therefore, the body's acceleration at t = 3 seconds is 6 m/s^2, and at t = 1 second is -6 m/s^2.
To find the body's acceleration each time the velocity is zero, we need to analyze the given position function and determine the velocity and acceleration functions.
Let's start by finding the velocity function. Velocity is the derivative of the position function with respect to time (t):
v(t) = s'(t)
To find s'(t), we take the derivative of the position function:
s'(t) = (t^3 - 6t^2 + 9t)' = 3t^2 - 12t + 9
Now that we have the velocity function, we can find the acceleration function. Acceleration is the derivative of the velocity function with respect to time (t):
a(t) = v'(t)
To find v'(t), we take the derivative of the velocity function:
v'(t) = (3t^2 - 12t + 9)' = 6t - 12
Finally, we are looking for the points in time (t) when the velocity is zero. This means we need to solve the equation v(t) = 0:
3t^2 - 12t + 9 = 0
We can factor out a common factor:
3(t^2 - 4t + 3) = 0
(t - 1)(t - 3) = 0
So, we have two solutions:
t - 1 = 0 --> t = 1
t - 3 = 0 --> t = 3
Therefore, when t = 1 sec and t = 3 sec, the body's velocity is zero.
Now, let's find the body's acceleration at these points.
Substituting t = 1 into the acceleration function:
a(t) = 6t - 12
a(1) = 6(1) - 12
a(1) = 6 - 12
a(1) = -6 m/s^2
So, the body's acceleration at t = 1 sec is -6 m/s^2.
Substituting t = 3 into the acceleration function:
a(t) = 6t - 12
a(3) = 6(3) - 12
a(3) = 18 - 12
a(3) = 6 m/s^2
Therefore, the body's acceleration at t = 3 sec is 6 m/s^2.
v = 3t^2-12t+9
= 3(t-3)(t-1)
a = 6t-12
so find a at t=1,3