To find the distance traveled by a car under constant acceleration, we can use the following kinematic equation:

$s = v_i t + \frac{1}{2} a t^2$

Where:

• $s$ is the distance traveled.
• $v_i$ is the initial velocity.
• $a$ is the acceleration.
• $t$ is the time.

However, because the time interval is given (from $t_i = 20$ seconds to $t_f = 60$ seconds), we should calculate the distance traveled at both $t_f$ and $t_i$ and then subtract to find the distance during this interval.

Step 1: Convert time into hours

Since velocity and acceleration are given in km/h and km/h², we need to convert the time from seconds to hours: $t_i = \frac{20}{3600} \text{ hours} \approx 0.00556 \text{ hours}$ $t_f = \frac{60}{3600} \text{ hours} \approx 0.01667 \text{ hours}$

Step 2: Calculate the distance at $t_f$

Using the formula: $s_f = v_i t_f + \frac{1}{2} a t_f^2$ $s_f = 30 \times 0.01667 + \frac{1}{2} \times 2 \times (0.01667)^2$ $s_f \approx 0.5001 + 0.00028 = 0.50038 \text{ km}$

Step 3: Calculate the distance at $t_i$

Similarly: $s_i = v_i t_i + \frac{1}{2} a t_i^2$ $s_i = 30 \times 0.00556 + \frac{1}{2} \times 2 \times (0.00556)^2$ $s_i \approx 0.1668 + 0.00003 = 0.16683 \text{ km}$

Step 4: Find the distance traveled between $t_i$ and $t_f$

$\Delta s = s_f - s_i = 0.50038 - 0.16683 = 0.33355 \text{ km}$

So, the car travels approximately 0.33355 km during the time interval from 20 seconds to 60 seconds.

Would you like more details or have any questions?

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Here are 5 related questions:

1. How would the distance change if the initial velocity were doubled?
2. What would be the distance if the acceleration were negative?
3. How does the time interval affect the distance traveled?
4. What if the car started from rest; how would the distance calculation differ?
5. How would you calculate the final velocity at 60 seconds?

Tip: When solving kinematic problems, ensure that units are consistent across all variables before performing calculations.