The driver of a car wishes to pass a truck that is travelling at a constant speed of 20.0 m/sInitially, the car is also travelling at 20.0 m/s. Initially, the vehicles are separated by 25.0 m. andthe car pulls back into the truck's lane after it is 25.0 m ahead of the truck. The car is5.0 m long, and the truck is 20.0 m long. The car's acceleration is a constant 0.600 m/s(a) How much time is required for the car in this process?(błWhat distance does the car travel during this ? (c) What is the final speed of the car ?
Answers
Answer:
Explanation:
The driver of a car wishes to pass a truck that is travelling at a constant speed of 20.0 m/sInitially, the car is also travelling at 20.0 m/s. Initially, the vehicles are separated by 25.0 m. andthe car pulls back into the truck's lane after it is 25.0 m ahead of the truck. The car is5.0 m long, and the truck is 20.0 m long. The car's acceleration is a constant 0.600 m/s
Distance to be covered by car
= Distance covered by truck + Car Behind truck + Truck Length + Car ahead Truck + Car Length
= Distance covered by truck + 25 + 20 + 25 + 5
= Distance Covered by truck + 75 m
Let say T mins is required
Distance Covered by truck = 20T
Distance to be covered by car = 20T + 75
Distance to be covered by car = S = ut + (1/2)at²
= 20T + (1/2)(0.6)T²
= 20T + 0.3T²
Equating both
20T + 0.3T² = 20T + 75
=> T² = 75/0.3
=> T² = 250
=> T = √250
=> T = 15.8 sec
time is required for the car in this process = 15.8 sec
V = u + at
=> V = 20 + (0.6)(15.8)
=> V = 29.48 m/s
final speed of the car = 29.48 m/s
Distance = S = ut + (1/2)at²
= 20*15.8 + (1/2)(0.6)250
= 316 + 75
= 391 m
Distance traveled = 391 m
Answer:
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