Answer:
t = 9.14 s
Explanation:
We first analyze the accelerating motion by applying first equation of motion:
Vf₁ = Vi₁ + a₁t₁
where,
Vf₁ = Final Speed of Car before turning off engine
Vi₁ = Initial Speed of Car = 0 m/s
a₁ = acceleration of car = 25 m/s²
t₁ = time taken in accelerating motion
Therefore,
Vf₁ = 25t₁ ---------- equation (1)
Now, we apply second equation of motion:
s₁ = Vi₁ t₁ + (1/2)a₁t₁²
where,
s₁ = distance covered during accelerating motion
Therefore,
s₁ = (0)t₁ + (1/2)(25)t₁²
s₁ = 12.5 t₁² ----------- equation (2)
Now, we analyze the decelerating motion by applying first equation of motion:
Vf₂ = Vi₂ + a₂t₂
where,
Vf₂ = Final Speed of Car = 0 m/s
Vi₂ = Initial Speed of Car after turning off engine
a₂ = deceleration of car = - 3 m/s²
t₂ = time taken in decelerating motion
Therefore,
Vi₂ = 3t₂ ---------- equation (3)
Now, we apply second equation of motion:
s₂ = Vi₂ t₂ + (1/2)a₂t₂²
where,
s₂ = distance covered during decelerating motion
Therefore,
s₂ = (Vi₂)t₂ + (1/2)(-3)t₂²
s₂ = Vi₂ t₂ - 1.5 t₂²
using equation (3):
s₂ = 3 t₂² - 1.5 t₂²
s₂ = 1.5 t₂² ------------ equation (4)
Now, we know that the Final Velocity of accelerating motion (Vf₁) is equal to the initial velocity of decelerating motion (Vi₂):
Vf₁ = Vi₂
using equation (1) and equation (3):
25 t₁ = 3 t₂
t₁ = 0.12 t₂ ------------ equation (5)
Also, we know that sum of the distances is 200 m:
s₁ + s₂ = 200
using equation (2) and equation (4):
12.5 t₁² + 1.5 t₂² = 200
using equation (5):
12.5 (0.12 t₂²) + 1.5 t₂² = 200
3 t₂² = 200
t₂² = 200/3
t₂ = 8.16 s
substitute this in equation (5):
t₁ = 0.12(8.16 s)
t₁ = 0.97 s
Hence, the minimum time required for this motion is:
t = t₁ + t₂ = 0.97 s + 8.16 s
<u>t = 9.14 s</u>
