After t seconds, a particle has covered a distance of S meters, where S=t^3-3t^2+9t+1
Determined the time when:
(i) Velocity is zero
(ii)Acceleration is zero
(b)Find its velocity and acceleration when t=2 seconds.
(c) What is the acceleration after 6 second s?
S=t^3-3t^2+9t+1
then the velocity is
v = ds/dt = 3 t^2 - 6 t + 9
solve that quadratic for v = 0
then
a = dv/dt = 6 t - 6
obviously a = 0 when t = 1
I think you can put t = 2 in the v equation and t = 6 in the a equation
Pls show how to get t when V is zero
(i) To determine when the velocity is zero, we need to find the value of t when the derivative of the distance equation with respect to time is zero.
So, let's differentiate S(t) with respect to t:
S'(t) = 3t^2 - 6t + 9
To find when the velocity is zero, we set S'(t) equal to zero:
3t^2 - 6t + 9 = 0
Unfortunately, this quadratic equation doesn't factor nicely, so we need to use the quadratic formula. The formula is:
t = (-b ± √(b^2 - 4ac))/(2a)
Applying the formula to our equation, we have:
t = (-(-6) ± √((-6)^2 - 4(3)(9)))/(2(3))
t = (6 ± √(36 - 108))/(6)
t = (6 ± √(-72))/(6)
Since we're dealing with square roots of negative numbers, there are no real solutions to this equation. Hence, there is no time when the velocity is zero.
(ii) To find when the acceleration is zero, we need to find the value of t when the derivative of the velocity equation with respect to time is zero.
So, let's differentiate V(t) with respect to t (where V(t) is the derivative of S(t)):
V'(t) = 6t - 6
To find when the acceleration is zero, we set V'(t) equal to zero:
6t - 6 = 0
t = 1
Hence, the time when the acceleration is zero is t = 1.
(b) To find the velocity and acceleration when t = 2 seconds, we substitute t = 2 into the appropriate equations:
S(2) = (2)^3 - 3(2)^2 + 9(2) + 1 = 15 meters (distance)
V(2) = 6(2) - 6 = 6 meters per second (velocity)
A(2) = 6 meters per second squared (acceleration)
Hence, when t = 2 seconds, the velocity is 6 m/s and the acceleration is also 6 m/s².
(c) To find the acceleration after 6 seconds, we substitute t = 6 into the acceleration equation:
A(6) = 6(6) - 6 = 30 meters per second squared.
So, after 6 seconds, the acceleration is 30 m/s².
To determine the time when velocity is zero, we need to find the value of t that makes the derivative of the distance equation S equal to zero. Let's start by finding the derivative of S with respect to t.
(i) Velocity is zero:
To find when the velocity is zero, we need to find the root of the derivative of the distance equation S.
1. Start with the equation for distance:
S(t) = t^3 - 3t^2 + 9t + 1
2. Differentiate the equation with respect to t:
dS(t)/dt = 3t^2 - 6t + 9
3. Set the derivative equal to zero and solve for t:
3t^2 - 6t + 9 = 0
This quadratic equation can be solved using the quadratic formula or factoring.
(ii) Acceleration is zero:
The acceleration is the derivative of velocity. To find when the acceleration is zero, we need to differentiate the derivative of the distance equation.
1. Start with the derivative of distance:
dS(t)/dt = 3t^2 - 6t + 9
2. Differentiate the equation again with respect to t to find acceleration:
d^2S(t)/dt^2 = 6t - 6
3. Set the second derivative equal to zero and solve for t:
6t - 6 = 0
Solve for t to find the time when acceleration is zero.
(b) To find the velocity and acceleration when t = 2 seconds, substitute t = 2 into the velocity and acceleration equations:
Velocity at t = 2 seconds:
v(2) = 3(2)^2 - 6(2) + 9
Acceleration at t = 2 seconds:
a(2) = 6(2) - 6
(c) To find the acceleration after 6 seconds, substitute t = 6 into the acceleration equation:
Acceleration at t = 6 seconds:
a(6) = 6(6) - 6
By following these steps and substituting the appropriate values, you can find the answers to the given questions.