The location x of a car in meters is given by the function x = 30t - 5t^2 where t is in seconds. At the time when the car is moving at 10 m/s in the direction opposite to its initial motion, how far is the car from where it was at t = 0? Show all your work.

Since the location x of the car in meters is given by the function x = 30t - 5t² where t is in seconds, we need to find the time at which its velocity is 10 m/s in the negative direction by differentiating x with respect to t to find its velocity, v.

So, v = dx/dt

= d(30t - 5t²)/dt

= d30t/dt - d5t²/dt

= 30 - 10t

When v is 10 m/s in the negative direction, v = -10 m/s.

So, v = 30 - 10t

-10 = 30 - 10t

-10 - 30 = -10t

-40 = -10t

t = -40/-10

t = 4 s

The distance at 4 s when its velocity is -10 m/s

Since at t = 4 s, its velocity is -10 m/s and x = 30t - 5t² is the car's location. The car's distance from t = 0 after its velocity is -10 m/s is

x(4) - x(0) = 30(4) - 5(4)² - [30(0) - 5(0)²]

= 120 - 5(16) - [0 - 0]

= 120 - 80 - 0

= 40 m

So, how far is the car from where it was at t = 0 is 40 m

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