8. If a child on a bicycle can accelerate at an average rate of [tex]0.53 \, \text{m/s}^2[/tex], how long would it take to increase the bicycle's velocity from [tex]0.68 \, \text{m/s}[/tex] to [tex]0.89 \, \text{m/s}[/tex]?



Answer :

To find out how long it would take for the child on the bicycle to increase their velocity from [tex]\(0.68 \, \text{m/s} \, [N]\)[/tex] to [tex]\(0.89 \, \text{m/s} \, [N]\)[/tex], given an average acceleration of [tex]\(0.53 \, \text{m/s}^2\)[/tex], we can use the following kinematic equation:

[tex]\[ a = \frac{{v_f - v_i}}{t} \][/tex]

Where:
- [tex]\(a\)[/tex] is the acceleration
- [tex]\(v_f\)[/tex] is the final velocity
- [tex]\(v_i\)[/tex] is the initial velocity
- [tex]\(t\)[/tex] is the time

We need to isolate [tex]\(t\)[/tex] (time) in the equation. Rearranging the equation to solve for [tex]\(t\)[/tex], we get:

[tex]\[ t = \frac{{v_f - v_i}}{a} \][/tex]

We can now substitute the given values into this equation:
- Initial velocity [tex]\(v_i = 0.68 \, \text{m/s}\)[/tex]
- Final velocity [tex]\(v_f = 0.89 \, \text{m/s}\)[/tex]
- Acceleration [tex]\(a = 0.53 \, \text{m/s}^2\)[/tex]

Substitute these values into the formula for time:

[tex]\[ t = \frac{{0.89 \, \text{m/s} - 0.68 \, \text{m/s}}}{0.53 \, \text{m/s}^2} \][/tex]

Next, calculate the difference in the velocities:

[tex]\[ 0.89 \, \text{m/s} - 0.68 \, \text{m/s} = 0.21 \, \text{m/s} \][/tex]

Then, divide this difference by the acceleration:

[tex]\[ t = \frac{0.21 \, \text{m/s}}{0.53 \, \text{m/s}^2} \][/tex]

Conducting this division gives:

[tex]\[ t \approx 0.39622641509433953 \, \text{s} \][/tex]

Therefore, it would take approximately [tex]\(0.396 \, \text{s}\)[/tex] for the child on the bicycle to accelerate from [tex]\(0.68 \, \text{m/s}\)[/tex] to [tex]\(0.89 \, \text{m/s}\)[/tex] at an average acceleration rate of [tex]\(0.53 \, \text{m/s}^2\)[/tex].