A speed bike tops a hill at 3.5 m/s and accelerates steadily down hill reaching a speed of 11.4 m/s after 4.2 seconds. How far did the bike travel during this period?
average speed=( 11.4+3.5 )/2
distance= avgspeed*time
a=(v-vₒ)/t
s= vₒ•t+at²/2
To find the distance traveled by the bike during this period, we can use the equation of motion:
\[ \text{Distance} = \text{Initial velocity} \times \text{time} + \frac{1}{2} \times \text{acceleration} \times \text{time}^2 \]
First, let's find the acceleration of the bike. We can use the formula for acceleration:
\[ \text{Acceleration} = \frac{\text{Final velocity} - \text{Initial velocity}}{\text{time}} \]
Substituting the given values:
\[ \text{Acceleration} = \frac{11.4 \, \text{m/s} - 3.5 \, \text{m/s}}{4.2 \, \text{s}} \]
Now we can substitute the acceleration into the distance equation:
\[ \text{Distance} = (3.5 \, \text{m/s}) \times (4.2 \, \text{s}) + \frac{1}{2} \times \left(\frac{11.4 \, \text{m/s} - 3.5 \, \text{m/s}}{4.2 \, \text{s}}\right) \times (4.2 \, \text{s})^2 \]
Simplifying this equation will give us the distance traveled by the bike during this period.