Question

a bicyclist, initially traveling 8m/s, decreases speed at a rate of 2m/s² for 4s. make a motion chart (from initial position) chart to answer the following questions. answers have 4% tolerance to allow for approximations. a = unit t unit v unit δx unit 0 1 2 3 4

Explanation:

Step1: Identify acceleration

The problem states the bicyclist decreases speed at a rate of $2m/s^{2}$. Since it's a decrease in speed, the acceleration $a=- 2m/s^{2}$.

Step2: Find velocity - time relation

The formula for velocity $v = v_0+at$, where $v_0 = 8m/s$ (initial velocity), $a=-2m/s^{2}$ and $t$ is time.

• At $t = 0s$, $v=v_0=8m/s$, $\Delta x = 0m$ (starting from initial position).
• At $t = 1s$, $v=v_0+at=8+( - 2)\times1=6m/s$. The displacement $\Delta x$ in the first - second is given by $\Delta x_1=v_0t+\frac{1}{2}at^{2}=8\times1+\frac{1}{2}\times(-2)\times1^{2}=8 - 1 = 7m$.
• At $t = 2s$, $v=v_0+at=8+( - 2)\times2=4m/s$. The displacement from the start $\Delta x_2=v_0t+\frac{1}{2}at^{2}=8\times2+\frac{1}{2}\times(-2)\times2^{2}=16 - 4 = 12m$.
• At $t = 3s$, $v=v_0+at=8+( - 2)\times3=2m/s$. The displacement from the start $\Delta x_3=v_0t+\frac{1}{2}at^{2}=8\times3+\frac{1}{2}\times(-2)\times3^{2}=24 - 9 = 15m$.
• At $t = 4s$, $v=v_0+at=8+( - 2)\times4=0m/s$. The displacement from the start $\Delta x_4=v_0t+\frac{1}{2}at^{2}=8\times4+\frac{1}{2}\times(-2)\times4^{2}=32 - 16 = 16m$.

Answer:

$a=-2$; unit: $m/s^{2}$

t (unit: s)	v (unit: m/s)	$\Delta x$ (unit: m)
1	6	7
2	4	12
3	2	15
4	0	16