A particle is moving with acceleration a(t)=6t+4. its position at time t=0 is s(0)=6 and its velocity at time t=0 is v(0)=16. What is its position at time t=13?

Question

anonymous · Answer

s = s_0 + v_0*t + a*t^2
where s_0 is the initial position, v_0 is the initial velocity, a is the acceleration, t is the change is time.

anonymous · Answer

you have to take the integral twice
$$\int\limits_{}^{}6t+4 = 3t ^{2}+4t+16$$
$$\int\limits_{}^{}3t ^{2}+4t+16= t ^{3}+2t ^{2}+16t+6$$

anonymous · Answer

then just plug in 13 for $$t ^{3}+2t ^{2}+16t+6$$

anonymous · Answer

makes sense. thanks man

radar · Answer

That looks like a good answer, does it agree with the "approved" answer?  Or do you have an approved answer?

radar · Answer

I know the integration of acceleration will produce the velocity.  then the integration of velocity will produce s or distance.  The integration looks good, I am going to give a good answer to jmoreno.

anonymous · Answer

it does agree with the "approved" answer. I wasn't aware that you needed the second integral