a jeep starts from rest with constant acc of 4 m/s^2. at the same time a car travels with a comstant speed of 36 km/hrs over takes and passes the jeep. how far beyond th starting point the jeep overtake the car?

Question

Saurabh Koranglekar · Answer

Khimraj · Answer

acceleration of jeep a = 4 m/s 2 speed of car v = 36 km/hr = 36*5/18 = 10m/s to overtake distance travelled by both should be same. so ½*4*t 2 = 10*t 2t = 10 t = 5 sec distance covered = 10*5 = 50m Hope it clears...................

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a jeep starts from rest with constant acc of 4 m/s^2. at the same time a car travels with a comstant speed of 36 km/hrs over takes and passes the jeep. how far beyond th starting point the jeep overtake the car?

a jeep starts from rest with constant acc of 4 m/s^2. at the same time a car travels with a comstant speed of 36 km/hrs
over takes and passes the jeep. how far beyond th starting point the jeep overtake the car?

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a jeep starts from rest with constant acc of 4 m/s^2. at the same time a car travels with a comstant speed of 36 km/hrs over takes and passes the jeep. how far beyond th starting point the jeep overtake the car?

a jeep starts from rest with constant acc of 4 m/s^2. at the same time a car travels with a comstant speed of 36 km/hrsover takes and passes the jeep. how far beyond th starting point the jeep overtake the car?

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a jeep starts from rest with constant acc of 4 m/s^2. at the same time a car travels with a comstant speed of 36 km/hrs
over takes and passes the jeep. how far beyond th starting point the jeep overtake the car?