the position function of a particle is given by

s=t^3–4t^2–7t t>=0

where s is measured in meters and t in seconds. When does the particle reach a velocity of 10 m/s?

Take the derivative:

v=3t^2 -8t -7

put v=10 and rearrange it to have a quadratic in standard format, solve the quadratic with the quadratic formula.

i got 1.39 and -4.06 but i still got it wrong

To find the time at which the particle reaches a velocity of 10 m/s, we need to differentiate the position function to obtain the velocity function and then solve for the time when the velocity is equal to 10 m/s.

1. Differentiate the position function with respect to time (t) to find the velocity function (v):
v = ds/dt

Given that:
s = t^3 – 4t^2 – 7t

Differentiating the above equation using the power rule, we get:
v = 3t^2 – 8t – 7

2. Set the velocity function equal to 10 and solve for t:
3t^2 – 8t – 7 = 10

Rearranging the equation:
3t^2 – 8t – 17 = 0

This is a quadratic equation. To solve it, we can either factor it or use the quadratic formula. In this case, we'll use the quadratic formula.

Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a

The values of a, b, and c in our equation are:
a = 3
b = -8
c = -17

Plugging these values into the quadratic formula:
t = (-(-8) ± √((-8)^2 - 4(3)(-17))) / (2(3))
t = (8 ± √(64 + 204)) / 6
t = (8 ± √268) / 6

Simplifying further, we have:
t = (8 ± √268) / 6

Now we have two possible values for t. Let's solve for both:

t1 = (8 + √268) / 6
t2 = (8 - √268) / 6

Calculating these values:

t1 ≈ 2.49 seconds (approximately)
t2 ≈ -1.16 seconds (approximately)

However, since time cannot be negative in this context, we discard the negative value.

Therefore, the particle reaches a velocity of 10 m/s at approximately t = 2.49 seconds.

To determine when the particle reaches a velocity of 10 m/s, we need to find the time t at which the particle's velocity is equal to 10 m/s.

The velocity of a particle is the derivative of its position function with respect to time. In this case, we need to find the derivative of the position function to find the velocity function.

Given the position function: s = t^3 - 4t^2 - 7t
To find the velocity function, take the derivative of the position function with respect to time (t):

v = ds/dt = d/dt (t^3 - 4t^2 - 7t)

Differentiating each term of the equation with respect to t:

v = 3t^2 - 8t - 7

Now we have the velocity function of the particle: v = 3t^2 - 8t - 7.

To find the time(s) when the velocity is equal to 10 m/s, set the velocity function equal to 10 and solve for t:

10 = 3t^2 - 8t - 7

Rearranging the equation:

3t^2 - 8t - 17 = 0

Now we have a quadratic equation. We can solve this equation using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values from our quadratic equation:

t = (-(-8) ± √((-8)^2 - 4(3)(-17))) / (2(3))

Simplifying:

t = (8 ± √(64 + 204)) / 6
t = (8 ± √268) / 6

Now, we evaluate the two possible values for t:

t1 = (8 + √ 268) / 6
t2 = (8 - √ 268) / 6

By solving for t1 and t2, we can determine when the particle reaches a velocity of 10 m/s.