From the image, the question involves calculating:

1. The acceleration of a bicyclist starting from rest and reaching a speed of $8.00 \, \text{m/s}$ in $4.60 \, \text{s}$.
2. The distance traveled during that time.

Step 1: Find Acceleration

The formula for acceleration ($a$) is: $a = \frac{\Delta v}{\Delta t}$ Where:

• $\Delta v = v_f - v_i = 8.00 \, \text{m/s} - 0.00 \, \text{m/s} = 8.00 \, \text{m/s}$
• $\Delta t = 4.60 \, \text{s}$

Substitute: $a = \frac{8.00}{4.60} = 1.74 \, \text{m/s}^2$

Step 2: Find Distance Traveled

The formula for distance ($d$) under constant acceleration is: $d = v_i t + \frac{1}{2} a t^2$ Where:

• $v_i = 0 \, \text{m/s}$ (initial velocity)
• $a = 1.74 \, \text{m/s}^2$
• $t = 4.60 \, \text{s}$

Substitute: $d = 0 + \frac{1}{2} (1.74)(4.60)^2$ $d = 0.87 (21.16)$ $d = 18.41 \, \text{m}$

Final Answers:

1. Acceleration: $1.74 \, \text{m/s}^2$
2. Distance: $18.41 \, \text{m}$

Would you like further clarification or a detailed breakdown of the steps?

5 Related Questions:

1. How does the formula for distance change if initial velocity is nonzero?
2. What is the relationship between acceleration and velocity for uniformly accelerated motion?
3. If the acceleration doubled, how would the distance traveled change?
4. Can this method be applied if the acceleration varies over time?
5. What is the equation for finding time when only distance and acceleration are given?

Tip:

Always check units during calculations to ensure consistency and accuracy.