The safe stopping distance, d, in metres, of a Harold Dobson motorcycle on wet pavement is given by the function d(s)= 0.02(3s^2 + 20s), where s is the speed of the motorcycle in metres per second. Find the speed at which the safe stopping distance is 50 metres.

0.02(3s^2+20s)=50

3s^2+20s=2500
3s^2+20s-2500=0
x=25.725992956937827
=25

set 0.02(3s^2 + 20s) = 50

3s^2 + 20s = 2500
3s^2 + 20s - 2500 = 0

Use the quadratic formula, reject the negative answer.

To find the speed at which the safe stopping distance is 50 meters, we need to set up and solve the equation d(s) = 50.

Given that the function for the safe stopping distance is d(s) = 0.02(3s^2 + 20s), we can substitute 50 for d(s):

50 = 0.02(3s^2 + 20s)

Now, let's simplify the equation:

50 = 0.06s^2 + 0.4s

Next, let's rearrange the equation to obtain a quadratic equation in standard form:

0.06s^2 + 0.4s - 50 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring may be a bit challenging, so let's use the quadratic formula:

s = (-b +/- √(b^2 - 4ac)) / 2a

In our equation, a = 0.06, b = 0.4, and c = -50. Substituting these values into the quadratic formula:

s = ( -0.4 +/- √(0.4^2 - 4 * 0.06 * -50)) / (2 * 0.06)

Now, let's calculate the values inside the square root:

s = ( -0.4 +/- √(0.16 + 12)) / 0.12

s = ( -0.4 +/- √(12.16)) / 0.12

s ≈ ( -0.4 +/- 3.48) / 0.12

This gives us two possible values for s. Let's calculate them separately:

s1 = ( -0.4 + 3.48) / 0.12 ≈ 28.93 m/s

s2 = ( -0.4 - 3.48) / 0.12 ≈ -37.32 m/s

Since speed cannot be negative in this context, we can disregard the negative value.

Therefore, the speed at which the safe stopping distance is 50 meters is approximately 28.93 m/s.

To find the speed at which the safe stopping distance is 50 meters, we need to set up an equation and solve for the speed, s.

The equation that represents the safe stopping distance is:
d(s) = 0.02(3s^2 + 20s)

We want to find the speed, s, when the safe stopping distance, d(s), is equal to 50 meters. Therefore, we can set up the equation as follows:

50 = 0.02(3s^2 + 20s)

To solve for s, we need to isolate it on one side of the equation. Let's start by dividing both sides of the equation by 0.02:

50/ 0.02 = (3s^2 + 20s)

Simplifying:

2500 = 3s^2 + 20s

Now, let's rearrange the equation to the standard quadratic form:
3s^2 + 20s - 2500 = 0

We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. The most straightforward method is to use the quadratic formula:

s = (-b ± √(b^2 - 4ac)) / (2a)

In our case, a = 3, b = 20, and c = -2500:

s = (-20 ± √(20^2 - 4*3*(-2500))) / (2*3)

Simplifying further:

s = (-20 ± √(400 + 30000)) / 6
s = (-20 ± √(30400)) / 6
s = (-20 ± 174.22) / 6

Now, we have two possible solutions:

s = (-20 + 174.22) / 6 = 154.22 / 6 ≈ 25.70 m/s

s = (-20 - 174.22) / 6 = -194.22 / 6 ≈ -32.37 m/s

Since speed cannot be negative in this context, we discard the negative solution. Therefore, the speed at which the safe stopping distance is 50 meters is approximately 25.70 m/s.