The position of a particle moving along a line is given by s(t)=2t^3=24t^2+90t+7 for t greater than or equal t

Question

The position of a particle moving along a line is given by s(t)=2t^3=24t^2+90t+7 for t greater than or equal to 0. For what values of t is the speed of the particles increasing?

Using speed instead of velocity is throwing me off, could you please do this step by step?

Thank you for your assistance!

Thank you for pointing that out s(t)=2t^3=24t^2+90t+7 is a type -> it should be s(t)=2t^3+24t^2+90t+7

Thank you for pointing that out s(t)=2t^3=24t^2+90t+7 is a typo -> it should be s(t)=2t^3+24t^2+90t+7

Answer 1

Speed is the absolute value of the velocity. If the velocity is negative, the particle will speed up if the velocity becomes "more negative", that is, if the velocity decreases, which means when the acceleration is negative. If the velocity is positive, the particle will speed up if the velocity becomes "more positive", that is, the velocity increases, which means the acceleration is positve. So a particle increases speed when velocity and acceleration are the same sign.

Answer 2

The velocity (!) of the particle is s'(t) = 6t^2 + 48t + 90, and the acceleration is 12t + 48. Acceleration is zero at t = -4. Velocity is zero at t = -7 and t = -8 (solving the quadratic). Since all negative times are disallowed, and there are no changes of sign for t > 0, the speed (or velocity) is increasing at all times.

Answer 3

v=ds/dt=6t^2-48t+90(I suppose s(t) =2tt^3-24t^2+..)and dv/dt=acceleration = 12t-48
the sign of acceleration is 0--------4+++++++ so for t>4 the speed is increasing and from 0 to 4 is decreasing. At t=0 the spead is 90 and at t=4 has a minimum of -6(reversed direction)which happened at t=3

Answer 4

Is this a typo: s(t)=2t^3=24t^2+90t+7
Do you mean s(t)=2t^3+24t^2+90t+7
or s(t)=2t^3-24t^2+90t+7