Question

19 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? a 9 m b 18 m c 27 m d 36 m

Explanation:

Step1: Identify the kinematic equation

We use the second - order kinematic equation for distance traveled under constant acceleration: $s = v_0t+\frac{1}{2}at^{2}$, where $v_0$ is the initial velocity, $t$ is the time, and $a$ is the acceleration.
Given: $v_0 = 6\ m/s$, $a = 2\ m/s^{2}$, $t = 3\ s$.

Step2: Substitute the values into the equation

Substitute $v_0 = 6$, $a = 2$, and $t = 3$ into the formula $s=v_0t + \frac{1}{2}at^{2}$.
First, calculate $v_0t$: $v_0t=6\times3 = 18$.
Then, calculate $\frac{1}{2}at^{2}$: $\frac{1}{2}\times2\times3^{2}=\frac{1}{2}\times2\times9 = 9$.
Now, add these two results together: $s=18 + 9=27$.

Answer:

C. 27 m