A 1400 sedan goes through a wide intersection traveling from north to south when

Question

A 1400 sedan goes through a wide intersection traveling from north to south when it is hit by a 2200 SUV traveling from east to west. The two cars become enmeshed due to the impact and slide as one thereafter. On-the-scene measurements show that the coefficient of kinetic friction between the tires of these cars and the pavement is 0.750, and the cars slide to a halt at a point 5.33 west and 6.49 south of the impact point How fast was sedan traveling just before the collision? How fast was SUV traveling just before the collision?

Explanation / Answer

The acceleration after impact is found from uR = ug(1400 + 2200) = (1400 + 2200)a a = ug = 0.75 x 9.8 = 7.35 The distance slid is SQRT(5.33^2 + 6.49^2) = 8.39 m Speed after impact from v^2 - u^2 = 2as v = 0 -u^2 = -2 x 7.35 x 8.39 u = 11.1 m/s and angle = tan^-1(6.49/5.33) = 50.42deg south of west. Resolving into south and west components and equating momentum before and after 1400v[1] = (1400 + 2200) x 11.1 x sin(50.42) 2200v[2] = (1400 + 2200) x 11.1 x cos(50.42) v[1] = 4.3 m/s v[2] =17.9.3 m/s

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