im trying to take the derivative of this problem and plug in seven but im not getting the answer.
s=t3–2t2–6t t> or equal to 0
where s is measured in meters and t in seconds. When does the particle reach a velocity of 7 m/s?
ds/dt = 3t^2 - 4t - 6
so when ds/dt = 7
3t^2 - 4t - 6 = 7
3t^2 - 4t - 13 = 0
Solve using the quadratic equation, and use the positive answer.
(I got 2.85 sec)
if I am driving a car at 100km / hr and my reaction time is 0.382 seconds.
with many thanks Stephen
To find the velocity of the particle, we need to take the derivative of the given equation for s with respect to t.
First, let's take the derivative of each term separately.
d/dt (t^3) = 3t^2
d/dt (2t^2) = 4t
d/dt (-6t) = -6
Now, we can put these derivatives together to find the derivative of s with respect to t.
s' = 3t^2 + 4t - 6
To find the time when the particle reaches a velocity of 7 m/s, we can set s' equal to 7 and solve for t.
7 = 3t^2 + 4t - 6
To solve this quadratic equation, we can rearrange it to form a standard quadratic equation:
3t^2 + 4t - 13 = 0
Now we can use the quadratic formula to find the values of t that satisfy this equation:
t = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 3, b = 4, and c = -13.
t = (-4 ± √(4^2 - 4 * 3 * -13)) / (2 * 3)
Simplifying this equation will give us two possible values for t.
t = (-4 ± √(16 + 156)) / (6) = (-4 ± √(172)) / (6)
Now we can simplify the expression further.
t = (-4 ± 2√(43)) / (6) = (-2 ± √(43)) / (3)
Therefore, the time (t) when the particle reaches a velocity of 7 m/s is approximately equal to (-2 + √(43)) / 3 or (-2 - √(43)) / 3.
To substitute the value of t as 7, we can evaluate (-2 + √(43)) / 3 and (-2 - √(43)) / 3 separately and compare them to see if they are approximately equal to 7.