A driver of a car travelling at 52 km h applies the brakes and accelerated uniformly in the opposite direction the car stops in the 5 second another driver going ahead 3 km h in another car applies is break slowly and stops in 10 seconds on the same graph paper plot the plot the speed virus time graph for the two cars which of the two cars travel further after the brakes were applied?
Answers
First Car A:
Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/s
Final velocity= V= 0m/s
time =t=5sec
Refer attachment for Graph :
Distance travelled : Area of traingle AOB =1/2 OB x AO
=(1/2)x 14.4 x 5=72/2=36m
Second Car B:
Initial Velocity= u= 3 km/hr=3x5/18=0.83m/s
Final velocity= V= 0m/s
time =t=10sec
Refer attachment for Graph :
Distance travelled : Area of traingle COD =1/2 OD x CO=(1/2)x 0.83 x 10=4.1m
So, after applying of brakes , Car A has travelled a distance of 3m and car B has travelled a distance of 4.1m.
∴ Car B has travelled farther than Car A after the brakes are applied.
Answer:
The speed v/s time graphs for the two cars can be plotted as follows.
The total displacement of each car can be obtained 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 = (1/2)*(10s)*(0.83ms-1) = 4.15 meters
Therefore, the first car is displaced by 36 meters whereas the second car
is displaced by 4.15 meters. Therefore, the first car (which was traveling
at 52 kmph) traveled farther post the application of brakes.