Hi:) anyone able to help with the last part? Thanks!:)

t=50s

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.