Answer:
t = 7.5 s
Explanation:
The distance traveled by the car at the time of meeting of the two cars must be the same. First, we calculate the distance traveled by the police car. For that we use 2nd equation of motion. Here, we take the time when police car starts to be reference. So,
s₁ = Vi t + (0.5)gt²
where,
s₁ = distance traveled by police car
Vi = Initial Velocity = 0 m/s
t = time taken
Therefore,
s₁ = (0 m/s)(t) + (0.5)(9.8 m/s²)t²
s₁ = 4.9 t²
Now, we calculate the distance traveled by the car. For constant speed and time to be 1 second more than the police car time, due to car starting time, we get:
s₂ = Vt = V(t + 1)
where,
s₂ = distance traveled by car
V = Velocity of car = (100 km/h)(1000 m/1 km)(1 h/ 3600 s) = 27.78 m/s
Therefore,
s₂ = 27.78 t + 27.78
Now, we know that at the time of meeting:
s₁ = s₂
4.9 t² = 27.78 t + 27.78
4.9 t² - 270.78 t - 27.78 = 0
solving the equation and choosing the positive root:
t = 6.5 s
since, we want to know the time from the moment car crossed police car. Therefore, we add 1 second of starting time in this.
t = 6.5 s + 1 s
t = 7.5 s
