Question

A particle moves in the xy plane with velocity v is equal to 80 - 3 y is equal to 2 if it passes through the point x is equal to 14 and y is equal to 4 at t is equal to 2 second find the equation of the path

Answer 1

Answer:

As we can see here v=(8t−2)i+2j

Therefore x component of velocity is dependent on time, thus it has acceleration in x direction, whereas it is a constant value 2 in y direction, therefore it has uniform velocity.

So this motion should be having a parabolic path.

for y component

y=2t+c

where c is a constant and 2 is velocity in y direction

for x component

x=ut+(8t2)/2+d

From 2nd equation of motion and 8 will be acceleration as it is the value of slope in v-t equation and d is constant.

Now at t=2,

x=14,y=4

Therefore putting values in above equation,

4=4+c

=>c=0

14=2u+(8∗4)/2+d

−2=2∗(−2)+d u is velocity of x at t=0

=>d=2

Therefore we get

y=2t

and

x=−2t+4t2+2

Eliminating t,

x=−y+y2+2

Therefore the trajectory of equation is

x=y2−y+2

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