Answer:
[tex]t=3.64s[/tex]
Step by step explanation
Step 1
In this step we use the definition of acceleration to determine the expression to integrate. Note that the acceleration is the derivative of velocity with respect to time.
[tex]a=\frac{dv}{dt} \\\implies adt=dv\\\implies dv=(32t^3-11t^4)dt[/tex]
Step 2
Perform integration on the expression from step 1. This calculation is performed as shown below.
[tex]v=\int(32t^3-11t^4)dt\\v=\frac{32}{4}t^4- \frac{11}{5}t^5 +c\\\8t^4- \frac{11}{5}t^5 +c[/tex]
Step 3
In this step we use the condition that at [tex]t=0[/tex], the velocity [tex]v=0[/tex] to find the exact form of the velocity function. We substitute this point into the velocity function we found in step 2.
[tex]0=8(0)^4-\frac{11}{5} (0)^5+c\\\implies c=0[/tex]
Step 4
We now use the function in step 3 to find out when the velocity is zero again.
[tex]v=0=8t^5-\frac{11}{5}t^4\\\implies 0=8t^4-\frac{11}{5} t^5\\\implies t^4(8-\frac{11}{5} t)\\\implies t=0, 8-\frac{11}{5}t=0\\\\\implies t=0,t=\frac{40}{11} =3.64[/tex]
The velocity is zero again when [tex]t=3.64[/tex]