Question

Roberto runs a 100-meter race. 7 seconds after the race started Roberto is 35 meters from the starting line and reaches his max speed; he runs at this max speed for the rest of the race. Roberto notices that he is 67 meters from the starting line 11 seconds after the race started. What is Roberto's max speed? 8 Correct meters per second Suppose Roberto runs for an additional z seconds after reaching his max speed... How far will Roberto travel during those additional z seconds? 8z Correct meters What is Roberto's distance from the starting line 7 + z seconds after the race started? 8(7+z) Incorrect meters What is Roberto's distance from the starting line x seconds after the race started (provided x ≥ 7 )? 8x+35 Incorrect meters

Answer 1

Answer:

$v_t=0.5\ m.s^{-1}$

$d_z=0.5z\ m$

$d_{7+z}=(65+0.5z)\ m$

$d_x=61.5+0.5x$

Explanation:

Given:

• total distance run by the racer, $d=100\ m$

• time taken by the racer to reach the top speed, $t=7\ s$

• distance run by the racer in 7 seconds, $d_7=100-35=65\ m$

• distance of the racer from the start after 11 seconds, $d_{11}=67\ m$

• 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:

$\Delta t=11-7=4\ s$

Distance covered in this time:

$\Delta d=67-65=2\ m$

Now the top speed:

$v_t=\frac{\Delta d}{\Delta t}$

$v_t=\frac{2}{4}$

$v_t=0.5\ m.s^{-1}$

The time racer runs at top speed $=z\ s$

Distance covered in this z seconds:

$d_z=v_t\times z$

$d_z=0.5z\ m$

Distance of the racer after (7+z) seconds:

$d_{z+7}=v_t.(7+z)$

$d_{7+z}=(65+0.5\times z)\ m$

$d_{7+z}=(65+0.5z)\ m$

Distance covered by the racer in x seconds where x>7:

$d_x=65+0.5\times (x-7)$

$d_x=65-3.5+0.5x$

$d_x=61.5+0.5x$