a driver of a car travelling at 52 km h minus 1 apply the breaker and acceleration uniformly in the opposite a direction the car stop in 5 second and other driver going at 3 km h n s 1 in antar apply this break slowly and stop in 10 second on the same graph paper plot the speed versus time graph for the same graph paper and two cars for the after the breakers
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Answer:
The total displacement of each car can be achieved by calculating the area beneath the speed-time graph.
Therefore, displacement of the first car = area of triangle
AOB= 1/2× (OB)×(OA)
But OB = 5 seconds and OA = 52 km.h-1-= 14.44 m/s
Therefore, the area of the triangle AOB is given by: 1/2×(5s)×(14.44ms-1) = 36 meters
Now, the displacement of the second car is given by the area of the triangle
COD= 1/2×(OD)×(OC)
But OC = 10 seconds and OC = 3km.h-1= 0.83 m/s
Therefore, area of triangle COD = 12× (10s)×(0.83ms-1) = 4.15 meters
Hence, the first car is displaced by 36 meters whereas the second car is displaced by 4.15 meters.
Therefore, the first car (which was travelling at 52 kmph) travelled farther post the application of brakes
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