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Sagot :
Alright, let's solve this problem step by step.
We are given:
- Initial velocity, [tex]\( u = 90 \, \text{km/h} \)[/tex]
- Final velocity, [tex]\( v = 54 \, \text{km/h} \)[/tex]
- Distance over which braking occurs, [tex]\( s = 40 \, \text{m} \)[/tex]
First, we need to convert the velocities from km/h to m/s.
1. Converting velocities:
[tex]\[ u = 90 \, \text{km/h} = 90 \times \frac{1000}{3600} \, \text{m/s} = 25 \, \text{m/s} \][/tex]
[tex]\[ v = 54 \, \text{km/h} = 54 \times \frac{1000}{3600} \, \text{m/s} = 15 \, \text{m/s} \][/tex]
2. Finding the deceleration [tex]\( a \)[/tex] using the equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Rearrange the equation to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{v^2 - u^2}{2s} \][/tex]
[tex]\[ a = \frac{(15)^2 - (25)^2}{2 \times 40} \][/tex]
[tex]\[ a = \frac{225 - 625}{80} \][/tex]
[tex]\[ a = \frac{-400}{80} \][/tex]
[tex]\[ a = -5 \, \text{m/s}^2 \][/tex]
The negative sign indicates deceleration.
3. Finding the time to slow down to [tex]\( 54 \, \text{km/h} \)[/tex]:
We use the equation:
[tex]\[ v = u + at \][/tex]
Rearrange to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{v - u}{a} \][/tex]
[tex]\[ t = \frac{15 - 25}{-5} \][/tex]
[tex]\[ t = \frac{-10}{-5} \][/tex]
[tex]\[ t = 2 \, \text{s} \][/tex]
Therefore, the time to slow down to 54 km/h is 2 seconds.
4. Finding the total time to come to rest:
When the bike comes to rest, the final velocity [tex]\( v = 0 \)[/tex]. We use the same equation [tex]\( v = u + at \)[/tex]:
[tex]\[ 0 = 25 + (a \times t) \][/tex]
[tex]\[ t = \frac{-u}{a} \][/tex]
[tex]\[ t = \frac{-25}{-5} \][/tex]
[tex]\[ t = 5 \, \text{s} \][/tex]
Thus, the total time to come to rest is 5 seconds.
5. Finding the total distance travelled until rest:
We use the equation [tex]\( v^2 = u^2 + 2as \)[/tex] where [tex]\( v = 0 \)[/tex] and solve for [tex]\( s \)[/tex]:
[tex]\[ 0 = u^2 + 2as \][/tex]
[tex]\[ s = \frac{-u^2}{2a} \][/tex]
[tex]\[ s = \frac{-(25)^2}{2 \times -5} \][/tex]
[tex]\[ s = \frac{625}{10} \][/tex]
[tex]\[ s = 62.5 \, \text{m} \][/tex]
Therefore, the total distance travelled by the bike until it comes to rest is 62.5 meters.
In conclusion:
(i) The total time in which the bike comes to rest is 5 seconds.
(ii) The total distance travelled by the bike is 62.5 meters.
We are given:
- Initial velocity, [tex]\( u = 90 \, \text{km/h} \)[/tex]
- Final velocity, [tex]\( v = 54 \, \text{km/h} \)[/tex]
- Distance over which braking occurs, [tex]\( s = 40 \, \text{m} \)[/tex]
First, we need to convert the velocities from km/h to m/s.
1. Converting velocities:
[tex]\[ u = 90 \, \text{km/h} = 90 \times \frac{1000}{3600} \, \text{m/s} = 25 \, \text{m/s} \][/tex]
[tex]\[ v = 54 \, \text{km/h} = 54 \times \frac{1000}{3600} \, \text{m/s} = 15 \, \text{m/s} \][/tex]
2. Finding the deceleration [tex]\( a \)[/tex] using the equation:
[tex]\[ v^2 = u^2 + 2as \][/tex]
Rearrange the equation to solve for [tex]\( a \)[/tex]:
[tex]\[ a = \frac{v^2 - u^2}{2s} \][/tex]
[tex]\[ a = \frac{(15)^2 - (25)^2}{2 \times 40} \][/tex]
[tex]\[ a = \frac{225 - 625}{80} \][/tex]
[tex]\[ a = \frac{-400}{80} \][/tex]
[tex]\[ a = -5 \, \text{m/s}^2 \][/tex]
The negative sign indicates deceleration.
3. Finding the time to slow down to [tex]\( 54 \, \text{km/h} \)[/tex]:
We use the equation:
[tex]\[ v = u + at \][/tex]
Rearrange to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{v - u}{a} \][/tex]
[tex]\[ t = \frac{15 - 25}{-5} \][/tex]
[tex]\[ t = \frac{-10}{-5} \][/tex]
[tex]\[ t = 2 \, \text{s} \][/tex]
Therefore, the time to slow down to 54 km/h is 2 seconds.
4. Finding the total time to come to rest:
When the bike comes to rest, the final velocity [tex]\( v = 0 \)[/tex]. We use the same equation [tex]\( v = u + at \)[/tex]:
[tex]\[ 0 = 25 + (a \times t) \][/tex]
[tex]\[ t = \frac{-u}{a} \][/tex]
[tex]\[ t = \frac{-25}{-5} \][/tex]
[tex]\[ t = 5 \, \text{s} \][/tex]
Thus, the total time to come to rest is 5 seconds.
5. Finding the total distance travelled until rest:
We use the equation [tex]\( v^2 = u^2 + 2as \)[/tex] where [tex]\( v = 0 \)[/tex] and solve for [tex]\( s \)[/tex]:
[tex]\[ 0 = u^2 + 2as \][/tex]
[tex]\[ s = \frac{-u^2}{2a} \][/tex]
[tex]\[ s = \frac{-(25)^2}{2 \times -5} \][/tex]
[tex]\[ s = \frac{625}{10} \][/tex]
[tex]\[ s = 62.5 \, \text{m} \][/tex]
Therefore, the total distance travelled by the bike until it comes to rest is 62.5 meters.
In conclusion:
(i) The total time in which the bike comes to rest is 5 seconds.
(ii) The total distance travelled by the bike is 62.5 meters.
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