a driver of a car travelling at 52 km/h applies the breaks and accelerates uniformly in the opposite direction the car stopped in 5s another driver going at 34 km/h in another car applies his breaks slowly and stops on 10s on the same graph plot the speed Vs time graph for the two cars which of the two cars travelled farther after the brakes were applied
Answers
Answer:
47m
explanation:
Here, initial speed of 1 st car, u=52kmh=52×1000m60×60s=14.4m/s
final speed, v=0, time taken, t=5s
Acceleration=final speed - initial speedtime taken=0.14.45=−2.88m/s2
Negative sign is for retardation.
Similarly, for second car, u=34kmh=34×1000m60×60s=9.4m/s, v=0, t=10s
Acceleration =v−ut=0−9.410=−0.94m/s2
Negative sign shows retardation.
To plot the speed-time graph, we take time along X-axis and speed along Y-axis. Take a point A on Y-axis to represent a speed of 14.4m/s and take a point B on X-axis to represent a time of 5second. Join A,B. This is the speed-time graph for the 1st car,
Again, take a point C on Y-axis to represent a speed of 9.4m/s and a point D on X-axis to represent a time of 10 second. Join C,D. This is the speed time graph for the second car,
As distance travelled = area ensolved between the speed-time graph and the time axis, therefore, distance travelled by the first car before it stopes, i.e., s1= area of ΔOAB
=OA×OB2=14.4×52=36m
Distance travelled by second car before it stops =area ofΔCOD
=OC×OD2=9.4×102=47m