well, just do the integral:
v(t) = ʃ (5/4 t)(t^2+1)^2 dt
= 5/4 ʃ t^3 + 2t^2 + t dt
The velocity v (in m/s) of an object moving with acceleration a (in m/s^2) is given by v =ʃ a dt, where t is the time (in seconds). Find a formula for v, if a =(5/4t)(t^2+1)^2.
4 answers
That'd be
5/4 ʃ t^5 + 2t^3 + t dt
5/4 ʃ t^5 + 2t^3 + t dt
mmmhh
v = ∫(5t/4)(t^2 + 1)^2 dt
= (5t/4)(((1/3)(1/(2t)) (t^2 + 1)^3 + c
= (5/24)(t^2 + 1)^3 + c
differentiate to get
((5/4)(t)(t^2 + 1)^2
v = ∫(5t/4)(t^2 + 1)^2 dt
= (5t/4)(((1/3)(1/(2t)) (t^2 + 1)^3 + c
= (5/24)(t^2 + 1)^3 + c
differentiate to get
((5/4)(t)(t^2 + 1)^2
Hmmm. I don't get your integral at all. I think you have tripped over a shortcut.
1 answer