Suppose you are driving a car. Let s (t) denote your position in feet at time t (and assume s (t) is differentiable). Consider an interval of time [ 0, 40 ] (in seconds). At t = 0, you put your foot on the pedal softly and keep the same amount of force on the pedal for 10 seconds. At the 10 second mark, you begin increasing the force of your foot gradually until t = 20, when you see a red light in the distance, so you gradually lift your foot from the pedal. At t = 30, you begin braking, and finally stop at the red light at t = 40. At what time t does the sign of the second derivative switch from positive to negative or from negative to positive?

Question

Francesca Jenkins

Physics

15 December, 15:45

Suppose you are driving a car. Let s (t) denote your position in feet at time t (and assume s (t) is differentiable). Consider an interval of time [ 0, 40 ] (in seconds). At t = 0, you put your foot on the pedal softly and keep the same amount of force on the pedal for 10 seconds. At the 10 second mark, you begin increasing the force of your foot gradually until t = 20, when you see a red light in the distance, so you gradually lift your foot from the pedal. At t = 30, you begin braking, and finally stop at the red light at t = 40. At what time t does the sign of the second derivative switch from positive to negative or from negative to positive?

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Answers (1)

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Roy Kent · Answer 1

at t=20 the second derivative switches from positive to negative

Explanation:

s (t) denotes the position in time

s' (t) the first derivative denotes the speed in time

s'' (t) the second derivative, the one we want denotes the acceleration

From T=0 to T=10 the acceleration is 0.

From T=10 to T=20 the acceleration is positive

at T=20 the acceleration switches from positive to negative

at T=30 the acceleration becomes even more negative

and at T=40 the acceleration becomes 0