Answer :
Sure! Let's solve this problem step-by-step:
1. Identify the given data:
- Initial velocity ([tex]\( u \)[/tex]): [tex]\( 0.68 \, \text{m/s} \)[/tex]
- Final velocity ([tex]\( v \)[/tex]): [tex]\( 0.89 \, \text{m/s} \)[/tex]
- Acceleration ([tex]\( a \)[/tex]): [tex]\( 0.53 \, \text{m/s}^2 \)[/tex]
2. Determine the formula to use:
To find the time ([tex]\( t \)[/tex]) required to increase the velocity, we can use the first equation of motion:
[tex]\[ v = u + at \][/tex]
Rearrange this equation to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{v - u}{a} \][/tex]
3. Substitute the known values into the equation:
- [tex]\( v = 0.89 \, \text{m/s} \)[/tex]
- [tex]\( u = 0.68 \, \text{m/s} \)[/tex]
- [tex]\( a = 0.53 \, \text{m/s}^2 \)[/tex]
[tex]\[ t = \frac{0.89 \, \text{m/s} - 0.68 \, \text{m/s}}{0.53 \, \text{m/s}^2} \][/tex]
4. Perform the calculation:
[tex]\[ t = \frac{0.21 \, \text{m/s}}{0.53 \, \text{m/s}^2} \][/tex]
Simplifying the division:
[tex]\[ t \approx 0.3962 \, \text{seconds} \][/tex]
Therefore, it would take approximately [tex]\( 0.3962 \)[/tex] seconds for the child to increase the bicycle's velocity from [tex]\( 0.68 \, \text{m/s} \)[/tex] to [tex]\( 0.89 \, \text{m/s} \)[/tex] at an average rate of [tex]\( 0.53 \, \text{m/s}^2 \)[/tex].
1. Identify the given data:
- Initial velocity ([tex]\( u \)[/tex]): [tex]\( 0.68 \, \text{m/s} \)[/tex]
- Final velocity ([tex]\( v \)[/tex]): [tex]\( 0.89 \, \text{m/s} \)[/tex]
- Acceleration ([tex]\( a \)[/tex]): [tex]\( 0.53 \, \text{m/s}^2 \)[/tex]
2. Determine the formula to use:
To find the time ([tex]\( t \)[/tex]) required to increase the velocity, we can use the first equation of motion:
[tex]\[ v = u + at \][/tex]
Rearrange this equation to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{v - u}{a} \][/tex]
3. Substitute the known values into the equation:
- [tex]\( v = 0.89 \, \text{m/s} \)[/tex]
- [tex]\( u = 0.68 \, \text{m/s} \)[/tex]
- [tex]\( a = 0.53 \, \text{m/s}^2 \)[/tex]
[tex]\[ t = \frac{0.89 \, \text{m/s} - 0.68 \, \text{m/s}}{0.53 \, \text{m/s}^2} \][/tex]
4. Perform the calculation:
[tex]\[ t = \frac{0.21 \, \text{m/s}}{0.53 \, \text{m/s}^2} \][/tex]
Simplifying the division:
[tex]\[ t \approx 0.3962 \, \text{seconds} \][/tex]
Therefore, it would take approximately [tex]\( 0.3962 \)[/tex] seconds for the child to increase the bicycle's velocity from [tex]\( 0.68 \, \text{m/s} \)[/tex] to [tex]\( 0.89 \, \text{m/s} \)[/tex] at an average rate of [tex]\( 0.53 \, \text{m/s}^2 \)[/tex].
