Question

Problem 2 A policemen on a motor bike is initially at rest at the internectin between two "trot" when he observes a car crossing the intersection while the light is still red and immediately necelerates to catch the car. The car has an initial velocity of 8,00 m/a and accelerates at 1.00 m/ The policemen, which is initially at rest, accelerates at a rate of 3.00m/ a) What in the distance travelled by the police in the first 5.00 seconda of ita motion? What in b) What is the velocity the police after traveling 06.0 meters from ita storting point? ( pointa) d) After how many seconds will the velocity of the two vehicles (police and ear) be the snme? the distance traveled by the car in the same time? (5 pointa) ow long W 6 points)

Answer 1

a) We know v = 4 + ab here initially police is at nest therefore 4 = 0 therefore V = 0 3+ 3 times 5 = 15 m/s therefore in 5 sec it travels 15 times 5 = 75 m b) given distanced = 65 meters V^2 = 4 ^2 + 2as therefore V = squareroot 6 times 65 = 19.74 m/s c) when police catching the thief S_police = let + 1/2 at ^2 S_cas = let + 1/2 at^2 therefore S_police = S_car 0 + 1/2 times 3t^2 = 8t + 1/2 times t^2 therefore t^2 = 8A therefore at E = 8 sec it catches the thief d) we know v = 4 + at therefore V_police = V_car therefore 8 +t = 0 + 3t therefore 2t = 8 t = 4 sec therefore at t = 4 sec the velocity of the both of them.