IIT JEE , AIEEE Forum - Physics , Mechanics

nrki99

the displacement of the particle moving in a straight line is described by the relation s= 6+12t-2t^2. here s is in meters and t in seconds. the distance travelled by the particle in first 5 seconds is :

1-20m

2-32m

3-24m

4-26m

truly

velocity = V = ds/dt = 12 - 4t

acceleration = dV/dt = - 4

V = 0 ,....,12=4t ====>>> t=3sec

distance moved in 3 sec S1= _[0]

^[3] V dt = 12 t - 2t^2 = 36 - 18 = 18m

now direction reverses and

S2 = ut + 1/2 at^2 = 0 + 1/2*4*2^2 = 8 m

total distance = S1 + S2 = 18 + 8 = 26 m ........so option 4

nrki99

thanx for sln i am giving u salute.

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