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 :)

# 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

## 1 Answers

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