Question

required information. consider a runner traveling with an initial velocity of 3.4 m/s in the positive direction accelerates at a constant rate of 1.9 m/s² for a time of 5 seconds. note: this is a multi - part question. once an answer is submitted, you will be unable to return to this part. what distance does the runner cover during this process? the distance traveled by the runner is \boxed{} m.

Explanation:

Step1: Recall kinematic distance formula

The formula for distance under constant acceleration is:
$$s = v_0 t + \frac{1}{2} a t^2$$
where $v_0=3.4\ \text{m/s}$, $a=1.9\ \text{m/s}^2$, $t=5\ \text{s}$.

Step2: Calculate first term ($v_0 t$)

$$v_0 t = 3.4 \times 5 = 17\ \text{m}$$

Step3: Calculate second term ($\frac{1}{2} a t^2$)

$$\frac{1}{2} \times 1.9 \times 5^2 = 0.5 \times 1.9 \times 25 = 23.75\ \text{m}$$

Step4: Sum terms for total distance

$$s = 17 + 23.75$$

Answer:

40.75