Question

a car starts from rest and accelerates uniformly at 3.0 m/s². a second car starts from rest 6.0 s later in the same point and accelerates uniformly at 5.0 m/s². how long does it take the second car to overtake the first car?

Explanation:

Step1: Define Variables and Equations

Let the first car (starting from rest) have acceleration \(a_1 = 3.0\ m/s^2\), initial velocity \(u_1 = 0\ m/s\). The second car starts 6.0 s later, so its time of motion is \(t - 6\) when the first car has been moving for \(t\) seconds. The second car has acceleration \(a_2 = 5.0\ m/s^2\) and initial velocity \(u_2 = 0\ m/s\) (starts from rest? Wait, no, wait the problem says "a second car starts from rest 6.0 s later in the same point and accelerates uniformly at 5.0 m/s²". Wait, first car: starts from rest, accelerates at 3.0 m/s². Second car: starts from rest 6.0 s later, accelerates at 5.0 m/s². We need to find when the second car overtakes the first, so their displacements are equal.

Displacement for first car: \(s_1 = u_1t + \frac{1}{2}a_1t^2 = 0 + \frac{1}{2}(3)t^2 = 1.5t^2\)

Displacement for second car: it starts at \(t = 6\), so its time of motion is \(t - 6\), so \(s_2 = u_2(t - 6) + \frac{1}{2}a_2(t - 6)^2 = 0 + \frac{1}{2}(5)(t - 6)^2 = 2.5(t - 6)^2\)

Set \(s_1 = s_2\):

\(1.5t^2 = 2.5(t - 6)^2\)

Step2: Expand and Simplify

Expand the right side: \(2.5(t^2 - 12t + 36) = 2.5t^2 - 30t + 90\)

So equation: \(1.5t^2 = 2.5t^2 - 30t + 90\)

Subtract \(1.5t^2\) from both sides: \(0 = t^2 - 30t + 90\)

Wait, wait, that can't be. Wait, maybe I misread the problem. Wait, the first car "starts from rest and accelerates uniformly at 3.0 m/s²". The second car "starts from rest 6.0 s later in the same point and accelerates uniformly at 5.0 m/s²". Wait, no, maybe the second car starts with some initial velocity? Wait, the original problem (after OCR correction): "A car starts from rest and accelerates uniformly at 3.0 m/s². A second car starts from rest 6.0 s later in the same point and accelerates uniformly at 5.0 m/s². How long does it take the second car to overtake the first car?"

Wait, no, the question is "How long does it take the second car to overtake the first car?" So let \(t\) be the time the second car travels until overtaking. Then the first car has traveled \(t + 6\) seconds.

So displacement of first car: \(s_1 = 0.5 \times 3 \times (t + 6)^2 = 1.5(t + 6)^2\)

Displacement of second car: \(s_2 = 0.5 \times 5 \times t^2 = 2.5t^2\)

Set equal: \(1.5(t + 6)^2 = 2.5t^2\)

Ah, that's the correct setup. I had the time reversed before.

So expand left side: \(1.5(t^2 + 12t + 36) = 1.5t^2 + 18t + 54\)

Set equal to right side: \(1.5t^2 + 18t + 54 = 2.5t^2\)

Subtract \(1.5t^2\) from both sides: \(18t + 54 = t^2\)

Rearrange: \(t^2 - 18t - 54 = 0\)

Now use quadratic formula: \(t = \frac{18 \pm \sqrt{18^2 + 4 \times 54}}{2}\)

Calculate discriminant: \(324 + 216 = 540\), \(\sqrt{540} \approx 23.24\)

So \(t = \frac{18 + 23.24}{2} \approx \frac{41.24}{2} \approx 20.62\) s (we take the positive root)

Wait, but let's check the setup again. The first car starts at time 0, accelerates at 3 m/s². The second car starts at time 6 s, accelerates at 5 m/s². Let \(t\) be the time after the second car starts (so total time for first car is \(t + 6\)). Then:

\(s_1 = 0.5 \times 3 \times (t + 6)^2\)

\(s_2 = 0.5 \times 5 \times t^2\)

Set \(s_1 = s_2\):

\(1.5(t^2 + 12t + 36) = 2.5t^2\)

\(1.5t^2 + 18t + 54 = 2.5t^2\)

\(t^2 - 18t - 54 = 0\)

Quadratic formula: \(t = \frac{18 \pm \sqrt{324 + 216}}{2} = \frac{18 \pm \sqrt{540}}{2} = \frac{18 \pm 6\sqrt{15}}{2} = 9 \pm 3\sqrt{15}\)

Since time can't be negative, \(t = 9 + 3\sqrt{15} \approx 9 + 11.62 = 20.62\) seconds.

Wait, but maybe I made a mistake in the problem statement. Let me re-express the problem:

First car: initial velocity \(u_1 = 0\), acceleration \(a_1 = 3\ m/s^2\), time of motion \(T\) when overtaken.

Second car: initial velocity \(u_2 = 0\), acceleration \(a_2 = 5\ m/s^2\), time of motion \(T - 6\) (since it starts 6 s later).

Displacement: \(s_1 = 0.5a_1T^2\), \(s_2 = 0.5a_2(T - 6)^2\)

Set equal: \(0.5 \times 3 \times T^2 = 0.5 \times 5 \times (T - 6)^2\)

Cancel 0.5: \(3T^2 = 5(T^2 - 12T + 36)\)

\(3T^2 = 5T^2 - 60T + 180\)

\(0 = 2T^2 - 60T + 180\)

Divide by 2: \(T^2 - 30T + 90 = 0\)

Quadratic formula: \(T = \frac{30 \pm \sqrt{900 - 360}}{2} = \frac{30 \pm \sqrt{540}}{2} = \frac{30 \pm 6\sqrt{15}}{2} = 15 \pm 3\sqrt{15}\)

Now, \(T\) must be greater than 6 s (since second car starts at 6 s), so \(T = 15 + 3\sqrt{15} \approx 15 + 11.62 = 26.62\) s. Then the time for the second car is \(T - 6 = 20.62\) s, which matches the previous result.

Ah, so my initial setup had \(t\) as the second car's time, so \(T = t + 6\), so when I set \(s_1 = s_2\) with \(T = t + 6\), I get the same equation. So the correct quadratic is \(T^2 - 30T + 90 = 0\) when \(T\) is the first car's time, or \(t^2 - 18t - 54 = 0\) when \(t\) is the second car's time. Wait, no, let's do it again carefully.

Let \(t\) = time elapsed since the first car starts.

First car: \(s_1 = 0.5 \times 3 \times t^2 = 1.5t^2\)

Second car: starts at \(t = 6\), so its time of motion is \(t - 6\) (for \(t \geq 6\)), so \(s_2 = 0.5 \times 5 \times (t - 6)^2 = 2.5(t - 6)^2\)

Set \(s_1 = s_2\):

\(1.5t^2 = 2.5(t - 6)^2\)

Divide both sides by 0.5: \(3t^2 = 5(t^2 - 12t + 36)\)

\(3t^2 = 5t^2 - 60t + 180\)

\(0 = 2t^2 - 60t + 180\)

Divide by 2: \(t^2 - 30t + 90 = 0\)

Quadratic formula: \(t = \frac{30 \pm \sqrt{900 - 360}}{2} = \frac{30 \pm \sqrt{540}}{2} = \frac{30 \pm 6\sqrt{15}}{2} = 15 \pm 3\sqrt{15}\)

Now, \(15 - 3\sqrt{15} \approx 15 - 11.62 = 3.38\) s, which is less than 6 s, so invalid (since second car hasn't started yet). So valid solution is \(t = 15 + 3\sqrt{15} \approx 26.62\) s. Then the time the second car takes is \(t - 6 = 26.62 - 6 = 20.62\) s.

So the time for the second car to overtake is approximately 20.6 seconds (or exactly \(9 + 3\sqrt{15}\) seconds, which is about 20.6 s).

Answer:

The second car takes approximately \(\boxed{20.6}\) seconds (or exactly \(9 + 3\sqrt{15}\) seconds) to overtake the first car.