Answer :
Answer:
The object's position at t = 2 is 49.
Step-by-step explanation:
To find the object's position at t = 2, we need to find the velocity and position equations.
The relationship between position, velocity and acceleration:
- Velocity is the integral of acceleration.
- Position is the integral of velocity.
Given:
[tex]a(t)=30t-2[/tex]
Therefore:
[tex]v(t)=\int {a(t)} \, dt[/tex]
[tex]=\int {30t-2} \, dt[/tex]
[tex]\displaystyle =30\left(\frac{1}{1+1} \right)t^{1+1}-2\left(\frac{1}{0+1}\right) t^{0+1}+c[/tex]
[tex]=15t^2-2t+c[/tex]
Given:
[tex]v(0)=6[/tex]
[tex]15(0)^2-2(0)+c=6[/tex]
[tex]c=6[/tex]
Therefore:
[tex]\bf v(t)=15t^2-2t+6[/tex]
[tex]s(t)=\int{v(t)} \, dt[/tex]
[tex]= \int {15t^2-2t+6} \, dt[/tex]
[tex]\displaystyle =15\left(\frac{1}{2+1} \right)t^{2+1}-2\left(\frac{1}{1+1}\right) t^{1+1}+6\left(\frac{1}{0+1}\right) t^{0+1}+c[/tex]
[tex]=5t^3-t^2+6t+c[/tex]
Given:
[tex]s(0)=1[/tex]
[tex]5(0)^3-(0)^2+6(0)+c=1[/tex]
[tex]c=1[/tex]
Therefore:
[tex]\bf s(t)=5t^3-t^2+6t+1[/tex]
The object's position at t = 2:
[tex]s(2)=5(2)^3-(2)^2+6(2)+1[/tex]
[tex]=\bf 49[/tex]
