# A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by?

x(t) = bt2 − ct3

, where b = 2.40 m/s2 and

c = 0.120 m/s3

. What is the instantaneous speed of the car at t = 20.0 sec ?

A) −48 m/s

B) 0

C) 48 m/s

D) 96 m/s

I know the answer is A, but can someone explain to me why

, where b = 2.40 m/s2 and

c = 0.120 m/s3

. What is the instantaneous speed of the car at t = 20.0 sec ?

A) −48 m/s

B) 0

C) 48 m/s

D) 96 m/s

I know the answer is A, but can someone explain to me why

Well it definitely isn't any of the other ones because they're all positive.

Try x(19.99)

Try x(20)

Try x(20.01)

The values for x keep decreasing as time goes on at 20 s. So velocity must be negative at that moment.

Anyway, instantaneous velocity is the derivative of position with respect to time.

v(t) = dx/dt = 2bt - 3ct^2

At t = 20.0 s

v(20) = 2(2.4 m/s^2)(20 s) - 3(0.12 m/s^3)(20 s)^2 = -48 m/s

However, SPEED is positive 48 m/s. Because speed is just the magnitude of velocity and doesn't care about direction, so it can't be negative. Not sure if that was a typo or what. Tread carefully :)

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