Alright, let's find the solutions for both parts of the question step-by-step.
### Part (a): Finding the velocity at time [tex]\( t \)[/tex]
1. Given acceleration function:
[tex]\[
a(t) = 2t + 2
\][/tex]
2. Velocity is the integral of acceleration:
[tex]\[
v(t) = \int a(t) \, dt = \int (2t + 2) \, dt
\][/tex]
3. Compute the integral:
[tex]\[
\int (2t + 2) \, dt = \int 2t \, dt + \int 2 \, dt = t^2 + 2t + C
\][/tex]
Here, [tex]\( C \)[/tex] is the constant of integration.
4. Use the initial condition [tex]\( v(0) = 4 \)[/tex] to find [tex]\( C \)[/tex]:
[tex]\[
v(0) = 4 \\
\text{Substitute } t = 0 \text{ into the velocity function:} \\
v(0) = 0^2 + 2 \cdot 0 + C = C \\
\text{Therefore,} \\
C = 4
\][/tex]
5. Now, substitute [tex]\( C \)[/tex] back into the velocity function:
[tex]\[
v(t) = t^2 + 2t + 4
\][/tex]
So, the velocity at time [tex]\( t \)[/tex] is:
[tex]\[
v(t) = t^2 + 2t + 4
\][/tex]
### Part (b): Finding the distance traveled during the first 5 seconds
1. Position (or distance) is the integral of velocity:
[tex]\[
x(t) = \int v(t) \, dt = \int (t^2 + 2t + 4) \, dt
\][/tex]
2. Compute the integral:
[tex]\[
\int (t^2 + 2t + 4) \, dt = \int t^2 \, dt + \int 2t \, dt + \int 4 \, dt \\
= \frac{t^3}{3} + t^2 + 4t + C'
\][/tex]
Here, [tex]\( C' \)[/tex] is another constant of integration. However, we are interested in the displacement between two points in time, from [tex]\( t = 0 \)[/tex] to [tex]\( t = 5 \)[/tex], and the constant [tex]\( C' \)[/tex] will cancel out in this case.
3. Calculate the distance traveled from [tex]\( t = 0 \)[/tex] to [tex]\( t = 5 \)[/tex]:
[tex]\[
d = x(5) - x(0)
\][/tex]
4. Compute [tex]\( x(5) \)[/tex] and [tex]\( x(0) \)[/tex]:
[tex]\[
x(5) = \frac{5^3}{3} + 5^2 + 4 \cdot 5 = \frac{125}{3} + 25 + 20 = \frac{125}{3} + 45 = \frac{125 + 135}{3} = \frac{260}{3}
\][/tex]
[tex]\[
x(0) = \frac{0^3}{3} + 0^2 + 4 \cdot 0 = 0
\][/tex]
5. Distance traveled:
[tex]\[
d = x(5) - x(0) = \frac{260}{3} - 0 = \frac{260}{3}
\][/tex]
Thus, the distance traveled during the first 5 seconds is:
[tex]\[
\frac{260}{3} \text{ meters}
\][/tex]
These are the final results:
- Velocity at time [tex]\( t \)[/tex]: [tex]\( v(t) = t^2 + 2t + 4 \)[/tex]
- Distance traveled during the first 5 seconds: [tex]\( \frac{260}{3} \)[/tex] meters
