Question

a bicycle with an initial velocity of +6 m/s accelerates at a rate of +2 m/s² for 3 seconds. what distance does the bicycle travel during this time?
○ 36 m
○ 27 m
○ 18 m
○ 9 m

Explanation:

Step1: Recall kinematic distance formula

The formula for distance under constant acceleration is:
$$s = ut + \frac{1}{2}at^2$$
where $u$ = initial velocity, $a$ = acceleration, $t$ = time.

Step2: Substitute given values

Given $u=6\ \text{m/s}$, $a=2\ \text{m/s}^2$, $t=3\ \text{s}$:
$$s = (6 \times 3) + \frac{1}{2} \times 2 \times 3^2$$

Step3: Calculate each term

$$s = 18 + \frac{1}{2} \times 2 \times 9$$
$$s = 18 + 9$$

Step4: Sum terms for total distance

$$s = 27$$

Answer:

B. 27 m