On the Mumbai–Pune Expressway a car at \(100\,\text{km/h}\) comes up behind a truck at \(90\,\text{km/h}\). The car is fast, yet the overtaking seems to take for ever: the car creeps past the truck. A few minutes later a bus passes in the opposite direction. It is gone in a flash, although it is no faster than the truck.
A gap closes at the relative speed
Put yourself in the frame of the truck. In that frame the truck is at rest, and the car moves at \(100 - 90 = 10\,\text{km/h}\). The car must cover the whole gap, and the lengths of both vehicles, at this small relative speed. That is why the overtaking is slow.
In Fig 3.1 both cars move a long way in the frame of the road. The police car covers \(100\,\text{m}\) and the speeding car \(60\,\text{m}\). But the gap between them depends only on the difference. It closes at \(25 - 15 = 10\,\text{m/s}\), so a \(40\,\text{m}\) gap takes \(4\,\text{s}\). In the frame of the speeding car the picture is simpler still. That car stands still, and the police car covers \(40\,\text{m}\) at \(10\,\text{m/s}\).
A car at \(25\,\text{m/s}\) is \(150\,\text{m}\) behind a truck at \(15\,\text{m/s}\) on the same lane. After how many seconds does the car reach the truck?
A chase that starts from rest
A chase often starts with one body at rest and speeding up. The relative velocity then changes, but the relative acceleration is constant. So the equations of motion work in the frame of the body that moves steadily.
A car passes a parked police jeep at a steady \(15\,\text{m/s}\). At that instant the jeep starts from rest with an acceleration of \(2.5\,\text{m/s}^2\). When does the jeep catch the car? How far ahead is the car at the most?
Work in the frame of the car. At the start the jeep has \(u_{JC} = 0 - 15 = -15\,\text{m/s}\): it falls behind at first. Its relative acceleration is \(a_{JC} = 2.5 - 0 = 2.5\,\text{m/s}^2\).
The jeep catches the car when the relative displacement is zero again: \(\Delta x_{JC} = -15t + \tfrac{1}{2}(2.5)t^2 = 0\).
\(t(-15 + 1.25t) = 0\), so \(t = 12\,\text{s}\). The gap is largest when the relative velocity is zero: \(-15 + 2.5t = 0\) gives \(t = 6\,\text{s}\), and the gap is \(15 \times 6 - \tfrac{1}{2}(2.5)(6^2) = 90 - 45 = 45\,\text{m}\).
The jeep catches the car after \(12\,\text{s}\), \(180\,\text{m}\) from the start, moving at \(30\,\text{m/s}\). The car is never more than \(45\,\text{m}\) ahead, at \(6\,\text{s}\).
A bus starts from rest at a stop with an acceleration of \(1\,\text{m/s}^2\). At that instant a cyclist passes it at a steady \(6\,\text{m/s}\). After how many seconds is the cyclist farthest ahead of the bus?
Head on, the speeds add
Now let the two bodies move towards each other. Take the direction of A as positive, so \(v_B\) is negative. Then \(v_{AB} = v_A - v_B\) is the sum of the two speeds. The gap closes at this large rate. This is the bus on the expressway that is gone in a flash.
In Fig 3.2 neither car is fast, but the gap disappears in \(4\,\text{s}\). Choose "from car B". Car B stands still, and car A comes at \(25\,\text{m/s}\), faster than either car moves on the road. This is why a head-on crash is so much worse than a crash from behind. The damage depends on the relative speed, and head on it is the sum.
A train leaves Delhi for Agra at \(80\,\text{km/h}\). At the same moment another train leaves Agra for Delhi on the parallel track at \(70\,\text{km/h}\). The two stations are \(200\,\text{km}\) apart. When and where do the trains meet?
The trains move towards each other, so the gap closes at \(80 + 70 = 150\,\text{km/h}\).
The time to meet is \(200 / 150 = 4/3\,\text{h}\), which is \(80\,\text{min}\).
In that time the Delhi train covers \(80 \times 4/3 \approx 106.7\,\text{km}\). They meet about \(107\,\text{km}\) from Delhi and \(93\,\text{km}\) from Agra.
The faster train covers the larger share of the \(200\,\text{km}\), in the ratio of the speeds, \(80 : 70\).
Two cyclists are \(90\,\text{m}\) apart. Each rides at \(5\,\text{m/s}\). In case 1 they ride towards each other. In case 2 they ride in the same direction, one behind the other. What happens to the gap?
Each pair starts \(100\,\text{m}\) apart. Rank the pairs by how fast the gap closes, fastest first.
On a highway a car at \(25\,\text{m/s}\) overtakes a \(12\,\text{m}\) truck at \(20\,\text{m/s}\). The car is \(4\,\text{m}\) long, starts \(20\,\text{m}\) behind the truck and ends \(20\,\text{m}\) ahead of it. Roughly how many seconds does the overtaking take?
- Dividing the gap by the speed of one body.A gap closes at the relative speed. Divide by \(|v_A - v_B|\), not by \(v_A\).
- Adding the speeds when one body chases the other.A chase is the same direction, so the speeds subtract.
- Assuming the largest gap in a chase is at the start.If the chaser starts slower and speeds up, the gap grows until the two velocities are equal. Set the relative velocity to zero to find that instant.