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Q)A particle moves along the x-direction with constant acceleration.The displacement,measured from a convenient position,is 2m at time t=0 and is zero when t=10s.If the velocity of the particle is momentary zero when t=6s,determine the acceleration a and the velocity v when t=10s.PLZ SOLVE IT SIR.......................

Sibashis ghosh , 9 Years ago

Grade 11

2 Answers

atreyee

Last Activity: 9 Years ago

Let the initial velocity be u and acceleration be a. The displacement of the particle from t=0 to t=10s is -2m.

hence, 10u+50a=-2 and u+6a=0 (using the formulae ut+1/2at² = s and u+at=v)

Solving these two equations, we get a=0.2m/s² and u=-1.2m/s

at t=10s, the velocity v=-1.2+10*0.2=0.8m/s

Bandi Gayatri

Last Activity: 7 Years ago

Since the x-t graph is parabolic. Write the equation of a parabola. With vertex as (6,y). Find the equation of the parabola by substituting the values given. That is (10,0) & (0,2) . on getting the equation of the parabola, differentiate once and twice for velocity and acceleration respectively.

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