To solve the problem of determining the distance traveled by a car under constant acceleration, let's break down the steps using the information from the blackboard images and the problem statement.

Problem Breakdown:

• Given:
  • Constant acceleration $a = 2 \, \text{km/hr}^2$
  • Initial velocity $V_i = 30 \, \text{km/hr}$
  • Time frame: $t_i = 20 \, \text{seconds}$ to $t_f = 60 \, \text{seconds}$

Step 1: Convert Units

• Convert acceleration from $\text{km/hr}^2$ to $\text{m/s}^2$: $2 \, \text{km/hr}^2 = 2 \times \left(\frac{1000 \, \text{m}}{3600 \, \text{s}}\right)^2 = 2 \times \frac{1}{1296} \, \text{m/s}^2 = \frac{2}{1296} \, \text{m/s}^2 \approx 0.00154 \, \text{m/s}^2$

• Convert initial velocity from $\text{km/hr}$ to $\text{m/s}$: $V_i = 30 \, \text{km/hr} = 30 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{30000}{3600} \, \text{m/s} \approx 8.33 \, \text{m/s}$

Step 2: Use the Kinematic Equation

The distance traveled under constant acceleration can be found using the equation: $d = V_i t + \frac{1}{2} a t^2$

However, since we are given a time interval from $t_i = 20 \, \text{seconds}$ to $t_f = 60 \, \text{seconds}$, we first calculate the distance at $t = 60 \, \text{seconds}$ and subtract the distance at $t = 20 \, \text{seconds}$.

Step 3: Calculate Distance

1. Distance at $t = 60 \, \text{seconds}$: $d_f = V_i t_f + \frac{1}{2} a t_f^2 = 8.33 \times 60 + \frac{1}{2} \times 0.00154 \times 60^2$ $d_f = 499.8 \, \text{m} + 2.772 \, \text{m} \approx 502.572 \, \text{m}$

2. Distance at $t = 20 \, \text{seconds}$: $d_i = V_i t_i + \frac{1}{2} a t_i^2 = 8.33 \times 20 + \frac{1}{2} \times 0.00154 \times 20^2$ $d_i = 166.6 \, \text{m} + 0.308 \, \text{m} \approx 166.908 \, \text{m}$

3. Net Distance Traveled between $t = 20 \, \text{seconds}$ and $t = 60 \, \text{seconds}$: $\Delta d = d_f - d_i = 502.572 \, \text{m} - 166.908 \, \text{m} \approx 335.664 \, \text{m}$

Final Answer:

The distance traveled by the car between 20 seconds and 60 seconds is approximately 335.66 meters.

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Would you like further details or explanations?

Related Questions:

1. How would the distance change if the acceleration was $4 \, \text{km/hr}^2$ instead?
2. What is the significance of converting units in solving physics problems?
3. How would you calculate the final velocity of the car after 60 seconds?
4. How would the distance change if the initial velocity was zero?
5. Can you derive the kinematic equations from first principles?

Tip:

Always make sure to convert all units to a consistent system (e.g., SI units) before performing calculations in physics to avoid errors.