Answer:
t=50s
Step-by-step explanation:
For speed-time graphs, the distance traveled can be found by finding the area under the graph.
When the 2 cars meet, the distance travelled by both of them would be equal.
Let's find the distance travelled by car A at t=40s.
Area of traingle= ½( base)( height)
Distance by Car A at t=40s
= ½(40)(20)
= 400m
Distance by Car B at 40s
= ½(20)(15) +20(15)
= 150 +300
= 450m
Difference in distance= 50m
From t=40s onwards, the cars travels at a constant speed and the increase in their total distance would be at a constant rate.
For every 20s, car A would travel an additional 20(20)= 400m. And car B would travel an additional 15(20)= 300m.
Hence for every 20s, the distance between the 2 cars would decrease by 400 -300= 100m. Thus, for 50m, additional time needed (from t=40s)= ½(20)= 10s.
Since to closen the gap by 100m, 20s is needed.
100m ----- 20s
50m ----- 20 ÷2= 10s
Therefore the cars would meet again 10s after t=40s, which is at t=50s.
Let's check!
At t=50s,
Distance travelled by Car A
= ½(40)(20) +10(20)
= 400 +200
= 600m
Distance travelled by Car B
= ½(20)(15) + 30(15)
= 150 +450
= 600m
✓ Both travelled an equal distance of 600m at t=50s.