A driver of a car travelling at 52 kilometre per hour applies the brakes and accelerates uniformly in the opposite direction. the car stops in 5 seconds .another driver going at 3 kilometre per hour in another car applies is break slowly and stop in 10 seconds. on the same graph paper clock speed versus time graph for the two cars. which of the car travel father after the brakes are applied?
Answers
Answer:
Given :
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:
heya.....
heya.....Given :
heya.....Given : First Car A: Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/s
heya.....Given : First Car A: Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/sFinal velocity= V= 0m/s
heya.....Given : First Car A: Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/sFinal velocity= V= 0m/stime =t=5sec
heya.....Given : First Car A: Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/sFinal velocity= V= 0m/stime =t=5secRefer attachment for Graph :Distance travelled : Area of traingle AOB =1/2 OB x AO=(1/2)x 14.4 x 5=72/2=36m
heya.....Given : First Car A: Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/sFinal velocity= V= 0m/stime =t=5secRefer attachment for Graph :Distance travelled : Area of traingle AOB =1/2 OB x AO=(1/2)x 14.4 x 5=72/2=36mSecond Car B:Initial Velocity= u= 3 km/hr=3x5/18=0.83m/s
heya.....Given : First Car A: Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/sFinal velocity= V= 0m/stime =t=5secRefer attachment for Graph :Distance travelled : Area of traingle AOB =1/2 OB x AO=(1/2)x 14.4 x 5=72/2=36mSecond Car B:Initial Velocity= u= 3 km/hr=3x5/18=0.83m/sFinal velocity= V= 0m/stime =t=10sec
heya.....Given : First Car A: Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/sFinal velocity= V= 0m/stime =t=5secRefer attachment for Graph :Distance travelled : Area of traingle AOB =1/2 OB x AO=(1/2)x 14.4 x 5=72/2=36mSecond Car B:Initial Velocity= u= 3 km/hr=3x5/18=0.83m/sFinal velocity= V= 0m/stime =t=10secRefer attachment for Graph :Distance travelled : Area of traingle COD =1/2 OD x CO=(1/2)x 0.83 x 10=4.1m
heya.....Given : First Car A: Initial Velocity= u= 52 km/hr=52x5/18=14.4 m/sFinal velocity= V= 0m/stime =t=5secRefer attachment for Graph :Distance travelled : Area of traingle AOB =1/2 OB x AO=(1/2)x 14.4 x 5=72/2=36mSecond Car B:Initial Velocity= u= 3 km/hr=3x5/18=0.83m/sFinal velocity= V= 0m/stime =t=10secRefer attachment for Graph :Distance travelled : Area of traingle COD =1/2 OD x CO=(1/2)x 0.83 x 10=4.1mSo, 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.