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Physics 21 Online
OpenStudy (anonymous):

please help me solve this.. A body of mass 2 kg at O has an initial velocity of 3 ms^-1 along OE and it is subjected to a force of 4 Newton(N) in a direction perpendicular to OE . The distance of body from O after 4 sec will be a. 12 m b. 20 m

OpenStudy (anonymous):

the above answer is wrong. Split up the initial direction as x and the perpendicular direction as y Now velocity along x is constant since force is perpendicular. so distance along x = v_x*t=12m Velocity along y is not constant and is equal to 2t since acceleration = F/m = 4n/2kg=2m/s^2 Then the displacement = t^2= 16 Now use pythagorean theorem to compute total displacement d^2=12^2+16^2=20^2 so d=20

OpenStudy (anonymous):

thanks a lot..can u pls tell y v have 2 take t^2 & d^2..

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