The position function of a particle is given by s(t) = t^{3}-3t^{2}-7t, t\ge 0 where s is measured in meters and t in seconds.
When does the particle reach a velocity of 20 m/s?
PLEASE HELP. I found the derivative which is 3t^2-6t-7. Now do I set that equal to 20 or do I try to put 20 in for t?
You want v=20, so, just solve
3t^2-6t-7 = 20
If you put in t=20, that will give you the velocity after 20 seconds. C'mon - think about what the variables mean...
To find the time at which the particle reaches a velocity of 20 m/s, you need to set the derivative of the position function equal to 20. The derivative of the position function is the velocity function, so you are finding the time when the velocity is 20 m/s.
Given that the derivative is 3t^2 - 6t - 7, you can set it equal to 20 and solve for t:
3t^2 - 6t - 7 = 20
Now, rearrange the equation to bring all terms to one side:
3t^2 - 6t - 7 - 20 = 0
3t^2 - 6t - 27 = 0
Next, you can solve this quadratic equation for t. There are different ways to do this, such as factoring, using the quadratic formula, or completing the square. In this case, it appears that factoring will not be straightforward, so let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c:
t = (-(-6) ± sqrt((-6)^2 - 4(3)(-27))) / (2(3))
Simplifying further:
t = (6 ± sqrt(36 + 324)) / 6
t = (6 ± sqrt(360)) / 6
Now, simplify the square root:
t = (6 ± sqrt(36 * 10)) / 6
t = (6 ± 6 * sqrt(10)) / 6
Simplify further:
t = 1 ± sqrt(10)
Therefore, the particle reaches a velocity of 20 m/s at t = 1 + sqrt(10) or t = 1 - sqrt(10).
Remember to always check your solutions by plugging them back into the original equation to ensure they are valid answers.