A car is travelling with a speed of 36 km/h. The driver applied the brakes and retards the car uniformly. The car is stopped in 5 sec. Find (i) The acceleration of car and (ii) Distance before it stops after Appling breaks?
(i)Here given: Initial velocity u=36 km/hu=36 km/h Final velocity v=0v=0 [Because car stopped] Time taken to stop the car t=5 sec.t=5 sec. On using equation, v=u+atv=u+at 0=36+a×50=36+a×5 Or 5a=−365a=−36 Or a=−365=−7.2 m/s2a=−365=−7.2 m/s2 Here negative (−ve)(−ve) sign means deceleration. Thus, the retardation(Negative acceleration)(Negative acceleration) of the Car is 7.2 m/s2 (ii)Initial speed, u=36 km/h = 36×518 = 10 m/s Time, t=5 s Final velocity, v=0 m/s Let a and s be acceleration and total distance before stopping respectively. Step 2: Formula used v=u+at (First equation of motion) v2=u2+2as (Third equation of motion) Step 3: Finding the retardation From the equation of motion Or,0=10+a×5 After solving a = -2 m/s2 Hence, the retardation is 2 m/s2. Step 4: Finding the distance traveled From the equation of motion v2=u2+2as Or,0=102−2×2×s After solving s = 25 m
