Answer:
The runner's acceleration was 
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

Solving for a:

The runner speeds up from vo=5 m/s to vf=9 m/s in t=4 seconds, thus:


The runner's acceleration was 
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With the switch open, there's no current in the circuit, and therefore
no voltage drop across any of the dissipative elements (the resistor
or the battery's internal impedance). So the entire battery voltage
appears across the switch, and the voltmeter reads 12.0V .