53
Catch-up motion (practice)
Textbook: Part 2, lesson 53, pp. 43–45
GoalCatch-up problems: find time, speed and gap.
New words
gap · oraliqcatching up · quvib yetishthe one behind · orqadagithe one ahead · oldindagi
Explanation
In catch-up problems three quantities are linked: the gap S, the catch-up speed v₂ − v₁ and the time t. S = (v₂ − v₁) × t, t = S : (v₂ − v₁), v₂ = S : t + v₁. For example, if the gap is 30 km, catching up takes 5 h and the front car goes 55 km/h, then the back car must go 30 : 5 + 55 = 61 km/h. After some time the gap is S − (v₂ − v₁) × t.
Worked examples
Gap 24 km, 14 and 10 km/h: 24 : (14 − 10) = 6 h.
Gap 40 km, 70 and 60 km/h: after 3 h it is 40 − (70 − 60) × 3 = 10 km.
Class activity
“What speed do we need?”: children choose a gap and a time and calculate the speed needed.
Practice
1
The cyclist behind rides at 14 km/h and the one ahead at 10 km/h; the gap is 24 km. After how many hours does he catch up?
6
2
The gap is 30 km and the car ahead goes 55 km/h. At what speed in km/h must the car behind go to catch up in 5 h?
61
3
The gap is 40 km, the one behind goes 70 km/h and the one ahead 60 km/h. How many km is the gap after 3 h?
10
4
The gap is 20 km; the car behind goes 80 km/h and the one ahead 70 km/h. How many km does the rear car travel before it catches up?
160