A particle starts from a point with a velocity of 6m,s and moves with an acceleration of -2m/s^2. Show that after 6s the particle will be at the starting point.

Question

jiteshmeghwal9 · Answer

@UnkleRhaukus plz help :)

unklerhaukus · Answer

$$x(t)=x_0+{v_x}_0\cdot t+\tfrac12a_x\cdot t^2$$

jiteshmeghwal9 · Answer

is this the third equation of motion ?

jiteshmeghwal9 · Answer

@UnkleRhaukus

unklerhaukus · Answer

its an equation of motion, under constant acceleration

jiteshmeghwal9 · Answer

ok! how this equation can prove the given question ?

unklerhaukus · Answer

if the object starts at the origin ie x_0 = 0[m],
and the initial velocity is v_x_0 = 6[m/s]
and the acceleration is a_x = -2[m/s^2]

unklerhaukus · Answer

what do you get when you substitute t= 6[s] ?

jiteshmeghwal9 · Answer

$$x(t)=x_0+v_{x0}.t+\frac{1}{2}a_x.t^2$$$$x(6)=0+6m/s+\frac{1}{2}-2(6)^2$$

jiteshmeghwal9 · Answer

$$x(6)=6-36$$$$6x=-30$$$$x=-5$$

unklerhaukus · Answer

you missed t on the middle term

unklerhaukus · Answer

$$x(6)=0+6m/s\times\color{brown}{6s}+\frac{1}{2}(-2m/s^2)(6s)^2$$

jiteshmeghwal9 · Answer

oh! yes sorry :P

jiteshmeghwal9 · Answer

$$6x=0$$$$x=0$$

unklerhaukus · Answer

do you mean x(6)    

the position x,  as a a function of time;  where the time is 6 seconds

jiteshmeghwal9 · Answer

so it is x(6)=0 only ?

unklerhaukus · Answer

we could have chosen a different $x_0$ [ie  $x(0)$],  but the function would still return the same place it started after six seconds.

jiteshmeghwal9 · Answer

ok!

jiteshmeghwal9 · Answer

so now it is the proof right ??