A car accelerates from rest at a stop sign at a rate of 3.0 m/s^2 to a speed of 21.0 m/s, and then immediately begins to decelerate to a stop at the next stop sign at a rate of 4.0 m/s^2. How long did it take the car to travel from the first stop sign to the second stop sign? Show your work.

t = 12.25 s

Explanation:

First we calculate the time taken by the car during accelerating motion, we use 1st equation of motion:

Vf = Vi + at₁

where,

Vf = Final Velocity = 21 m/s

Vi = Initial Velocity = 0 m/s

a = acceleration = 3 m/s²

t₁ = time taken during acceleration = ?

Therefore,

21 m/s = 0 m/s + (3 m/s²)t₁

t₁ = (21 m/s)/(3 m/s²)

t₁ = 7 s

Now, we use the same equation to find out the time taken during deceleration motion, with following data:

Vf = Final Velocity = 0 m/s

Vi = Initial Velocity = 21 m/s

a = deceleration = -4 m/s²

t₂ = time taken during deceleration = ?

Therefore,

0 m/s = 21 m/s + (-4 m/s²)t₂

t₂ = (-21 m/s)/(-4 m/s²)

t₂ = 5.25 s

Therefore, the total time taken by the car to travel from first stop sign to second stop sign is:

t = t₁ + t₂

t = 7 s + 5.25 s

t = 12.25 s