Respuesta :
Answer:
[tex]v_t=0.5\ m.s^{-1}[/tex]
[tex]d_z=0.5z\ m[/tex]
[tex]d_{7+z}=(65+0.5z)\ m[/tex]
[tex]d_x=61.5+0.5x[/tex]
Explanation:
Given:
- total distance run by the racer, [tex]d=100\ m[/tex]
- time taken by the racer to reach the top speed, [tex]t=7\ s[/tex]
- distance run by the racer in 7 seconds, [tex]d_7=100-35=65\ m[/tex]
- distance of the racer from the start after 11 seconds, [tex]d_{11}=67\ m[/tex]
- As the question states that the racer reaches his maximum speed after 7 seconds and continues with that.
So, the racer is at his top speed at 11 seconds also.
Time elapsed after reaching the top speed:
[tex]\Delta t=11-7=4\ s[/tex]
Distance covered in this time:
[tex]\Delta d=67-65=2\ m[/tex]
Now the top speed:
[tex]v_t=\frac{\Delta d}{\Delta t}[/tex]
[tex]v_t=\frac{2}{4}[/tex]
[tex]v_t=0.5\ m.s^{-1}[/tex]
The time racer runs at top speed [tex]=z\ s[/tex]
Distance covered in this z seconds:
[tex]d_z=v_t\times z[/tex]
[tex]d_z=0.5z\ m[/tex]
Distance of the racer after (7+z) seconds:
[tex]d_{z+7}=v_t.(7+z)[/tex]
[tex]d_{7+z}=(65+0.5\times z)\ m[/tex]
[tex]d_{7+z}=(65+0.5z)\ m[/tex]
Distance covered by the racer in x seconds where x>7:
[tex]d_x=65+0.5\times (x-7)[/tex]
[tex]d_x=65-3.5+0.5x[/tex]
[tex]d_x=61.5+0.5x[/tex]
