Answer:
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
