Question

Two cars leave one after the other and travel with an acceleration of 4 m/s². Two minutes after the departure of the first the distance between the cars becomes 28 km. The time interval between the departures of the cars is

• A. 190 sec
• B. 100 sec
• C. 80 sec
• D. 140 sec

Answer 1

Answer: (B)

100 sec

Answer 2

The problem involves two cars accelerating uniformly from rest. We need to find the time interval between their departures given their acceleration, the time elapsed since the first car departed, and the distance between them at that time.

1. Define Variables and Given Values:

• Acceleration of both cars, a = 4 m/s².
• Time elapsed since the first car departed, T = 2 minutes = 120 seconds.
• Distance between the cars at time T, Δx = 28 km = 28000 m.
• Initial velocity of both cars, u = 0 m/s (since they leave one after the other, implying starting from rest).
• Let Δt be the time interval between the departures of the cars (the unknown).

2. Equations of Motion: The displacement x of an object starting from rest with constant acceleration a after time t is given by: x = (1/2)at²

3. Displacement of Car 1: Car 1 starts at t = 0. At time T = 120 s, its displacement x₁ is: x₁ = (1/2)aT² x₁ = (1/2)(4)(120)² x₁ = 2 * 14400 x₁ = 28800 m

4. Displacement of Car 2: Car 2 starts Δt seconds after Car 1. So, at time T = 120 s (measured from Car 1's departure), Car 2 has been traveling for a time t₂ = T - Δt. For Car 2 to have traveled any distance, t₂ must be positive, meaning T - Δt > 0, or Δt < T. The displacement x₂ of Car 2 is: x₂ = (1/2)a(T - Δt)²

5. Distance Between the Cars: The distance between the cars Δx is the difference between their displacements (assuming they travel in the same direction from the same origin): Δx = x₁ - x₂ Δx = (1/2)aT² - (1/2)a(T - Δt)² Δx = (1/2)a [T² - (T - Δt)²] Using the difference of squares formula A² - B² = (A - B)(A + B) where A = T and B = (T - Δt): Δx = (1/2)a [(T - (T - Δt))(T + (T - Δt))] Δx = (1/2)a [Δt (2T - Δt)] Δx = (1/2)a [2TΔt - (Δt)²]

6. Substitute Values and Solve for Δt: Substitute the known values into the equation: 28000 = (1/2)(4) [2(120)Δt - (Δt)²] 28000 = 2 [240Δt - (Δt)²] Divide both sides by 2: 14000 = 240Δt - (Δt)² Rearrange into a standard quadratic equation form (Δt)² - 240Δt + 14000 = 0.

Use the quadratic formula Δt = [-b ± sqrt(b² - 4ac)] / 2a: Here, a = 1, b = -240, c = 14000. Δt = [240 ± sqrt((-240)² - 4 * 1 * 14000)] / (2 * 1) Δt = [240 ± sqrt(57600 - 56000)] / 2 Δt = [240 ± sqrt(1600)] / 2 Δt = [240 ± 40] / 2

This gives two possible solutions for Δt:

1. Δt₁ = (240 + 40) / 2 = 280 / 2 = 140 s
2. Δt₂ = (240 - 40) / 2 = 200 / 2 = 100 s

7. Physical Interpretation of Solutions: We established that for Car 2 to have covered any distance, Δt must be less than T (Δt < 120 s).

• Case 1: Δt = 140 s Since 140 s > 120 s, this means Car 2 would not have even started by the time T = 120 s. In this scenario, x₂ would be 0, and the distance between the cars would simply be x₁. Δx = x₁ = 28800 m = 28.8 km. However, the problem states Δx = 28 km. This contradiction means Δt = 140 s is not the physically correct solution.

• Case 2: Δt = 100 s Since 100 s < 120 s, this is a physically valid scenario. Let's verify this solution: x₁ = 28800 m Time Car 2 has been traveling, t₂ = T - Δt = 120 - 100 = 20 s. x₂ = (1/2)(4)(20)² = 2 * 400 = 800 m. The distance between the cars Δx = x₁ - x₂ = 28800 - 800 = 28000 m = 28 km. This matches the given information perfectly.

Therefore, the time interval between the departures of the cars is 100 seconds.

The final answer is $\boxed{\text{100 sec}}$

Solution:

The displacement of an object starting from rest with constant acceleration a after time t is given by $x = \frac{1}{2}at^2$.

Let $T$ be the time elapsed since the first car departed ($T = 2 \text{ minutes} = 120 \text{ s}$). Let $\Delta t$ be the time interval between the departures of the cars. The acceleration of both cars is $a = 4 \text{ m/s}^2$. The distance between the cars at time $T$ is $\Delta x = 28 \text{ km} = 28000 \text{ m}$.

1. Displacement of Car 1 ($x_1$): Car 1 travels for time $T$. $x_1 = \frac{1}{2}aT^2 = \frac{1}{2}(4)(120)^2 = 2(14400) = 28800 \text{ m}$.

2. Displacement of Car 2 ($x_2$): Car 2 starts $\Delta t$ seconds after Car 1. So, at time $T$, Car 2 has been traveling for $t_2 = T - \Delta t$. $x_2 = \frac{1}{2}a(T - \Delta t)^2$.

3. Distance between cars ($\Delta x$): $\Delta x = x_1 - x_2$ $\Delta x = \frac{1}{2}aT^2 - \frac{1}{2}a(T - \Delta t)^2$ $\Delta x = \frac{1}{2}a [T^2 - (T - \Delta t)^2]$ $\Delta x = \frac{1}{2}a [T^2 - (T^2 - 2T\Delta t + (\Delta t)^2)]$ $\Delta x = \frac{1}{2}a [2T\Delta t - (\Delta t)^2]$

4. Substitute values and solve for $\Delta t$: $28000 = \frac{1}{2}(4) [2(120)\Delta t - (\Delta t)^2]$ $28000 = 2 [240\Delta t - (\Delta t)^2]$ $14000 = 240\Delta t - (\Delta t)^2$ Rearrange into a quadratic equation: $(\Delta t)^2 - 240\Delta t + 14000 = 0$

   Using the quadratic formula $\Delta t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $\Delta t = \frac{240 \pm \sqrt{(-240)^2 - 4(1)(14000)}}{2(1)}$ $\Delta t = \frac{240 \pm \sqrt{57600 - 56000}}{2}$ $\Delta t = \frac{240 \pm \sqrt{1600}}{2}$ $\Delta t = \frac{240 \pm 40}{2}$

   Two possible solutions: $\Delta t_1 = \frac{240 + 40}{2} = \frac{280}{2} = 140 \text{ s}$ $\Delta t_2 = \frac{240 - 40}{2} = \frac{200}{2} = 100 \text{ s}$

5. Physical Validity Check: The time $T$ (120 s) is measured from the departure of the first car. For the second car to have covered any distance, it must have started before or at time $T$. Thus, $\Delta t$ must be less than $T$ ($\Delta t < 120 \text{ s}$).

   • If $\Delta t = 140 \text{ s}$, then $\Delta t > T$. This implies Car 2 had not yet started when the measurement was taken. In this case, $x_2 = 0$, and $\Delta x = x_1 = 28800 \text{ m} = 28.8 \text{ km}$. This contradicts the given $\Delta x = 28 \text{ km}$. So, $140 \text{ s}$ is not a valid physical solution.

   • If $\Delta t = 100 \text{ s}$, then $\Delta t < T$. This is physically valid. Let's verify: $x_1 = 28800 \text{ m}$ Car 2 travels for $t_2 = 120 - 100 = 20 \text{ s}$. $x_2 = \frac{1}{2}(4)(20)^2 = 2(400) = 800 \text{ m}$. $\Delta x = x_1 - x_2 = 28800 - 800 = 28000 \text{ m} = 28 \text{ km}$. This matches the given information.

Therefore, the time interval between the departures of the cars is 100 seconds.

The final answer is $\boxed{\text{100 sec}}$.