Final answer:
The drag racer takes 10.0 seconds to travel the distance of 475 meters, accelerating from rest at 9.1 m/s², resulting in a final speed of 91 m/s.
Step-by-step explanation:
The student has asked how long it would take for a drag racer to travel a distance of 475 meters starting from rest and accelerating at 9.1 m/s², as well as the speed at the end of the run. To solve this, we will use two kinematic equations:
First, the time t can be found using the formula:
s = ut + \frac{1}{2}at^2
Where:
- s is the distance (475 m)
- u is the initial velocity (0 m/s, since the car starts from rest)
- a is the acceleration (9.1 m/s²)
- t is the time
Rearranging and solving for t gives:
t = \sqrt{\frac{2s}{a}}
Plugging in the values:
t = \sqrt{\frac{2 \times 475 m}{9.1 m/s²}} = 10.0 seconds
Secondly, the final velocity v can be found using the formula:
v = u + at
Since u = 0:
v = 9.1 m/s² \times 10.0 s = 91 m/s
Therefore, the correct answer to the student's question is (a) 10.0 seconds, 91 m/s.