# The velocity of a car accelerating from rest increases to 120 km/h over a period of 30 sec. Estimate the distance traveled during this period. The velocity of a car that starts from rest increases to 120 km/h in 30 seconds. The velocity of the car in terms of m/s is 120*1000/3600 = 100/3 m/s. The acceleration of the car during the 30 seconds time interval is (100/3 - 0)/30 = 100/90 = 10/9 m/s^2.

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The velocity of a car that starts from rest increases to 120 km/h in 30 seconds. The velocity of the car in terms of m/s is 120*1000/3600 = 100/3 m/s. The acceleration of the car during the 30 seconds time interval is (100/3 - 0)/30 = 100/90 = 10/9 m/s^2.

The distance traveled by the car while its speed increases from 0 to 100/3 is given by D = u*t + (1/2)*a*t^2 where u is the initial velocity which in this case is 0. t is the time which is 30 seconds and a is the acceleration which is 10/9 m/s^2.

The distance traveled by the car is (1/2)*(10/9)*30 = 16.67 m

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