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.
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
That'd be
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
Hmmm. I don't get your integral at all. I think you have tripped over a shortcut.
To find the formula for v, we need to perform the integration of the given acceleration function a with respect to time t.
First, let's simplify the expression for a:
a = (5/4t)(t^2+1)^2
To simplify this, we can expand the square term:
a = (5/4t)(t^4 + 2t^2 + 1)
Now we can perform the integration:
v = ʃ a dt
v = ʃ [(5/4t)(t^4 + 2t^2 + 1)] dt
To integrate this, we can distribute the (5/4t) term and split the integral into three parts:
v = ʃ [(5/4t)(t^4)] dt + ʃ [(5/4t)(2t^2)] dt + ʃ [(5/4t)(1)] dt
Now, let's integrate each part:
v = (5/4) ʃ t^3 dt + (5/4) ʃ 2t dt + (5/4) ʃ dt
Integrating each part:
v = (5/4) * (1/4) t^4 + (5/4) * t^2 + (5/4) * t + C
Simplifying:
v = (5/16) t^4 + (5/4) t^2 + (5/4) t + C
So, the formula for v in terms of time t is:
v = (5/16) t^4 + (5/4) t^2 + (5/4) t + C
Here, C represents the constant of integration which could be determined if any initial conditions (i.e., the initial value or velocity) are provided.