pushkar raja
Last Activity: 7 Years ago
The product average velocity and time is the distance travelled by the car = 3.2m/s *10s = 32 m.
The initial velocity of the car = u = 0, uniform acceleration a = 2m/s^2. Then the final velocity = u+at = 0+2t = t.The deceleration being 2m/s^2, the time required to attain the zero velocity from the initial velocity of 2t is t. Tundefinedore the time of uniform running of the car = 10-2t seconds.
The distance travelled by the car from rest or the initial velocity of 0 m/s to reach the final velocity of 2t with a uniform acceleration of 2m/s^2 = (initial speed+final speed)/2 * time = 0+2t)t= t^2.
The car travels now with uniform speed of 2t m/s for (10-2t) secs. So the distance travelled = 2t(10-2t = 20t-4t^2.
The car now decelerates uniformly at 2m/s^2 and from the initial speed of 2tm/2 till its speed is zero, for this tha car takes t time. Therfore the distance travelled by the car in this time = (initial velocity+final velocity)/2 * time = {(2t+0)/2}*t = t^2.
Thus the distance travelled by the car = t^2+20t-4t^2 +t^2 which should be 32 meter.
Therefore 20t-2t^2 = 32. Or 2t^2-20t+32= 0. Or dividing this equation by 2 , we get: t^-10t +16 = 0. Or (t-8)(t-2) = 0 Therefore t= 2 is the possible time.
The time during which the car travels with urniform speed is 10-2t = 10-2*2 = 6 seconds. The uniform speed of the car = 2t = 2*2 = 4m/s. Distance travelled with uniform speed = 4*6 = 32 meter.
