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Sagot :
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
### 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
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