A and B practice the 100m race. They start from the starting point at the same time. A's speed is 7m / s, B's speed is 6.5m/s. After running for 10s, a sprained his foot, If the velocity is reduced to 6m / s, can b catch up with a before reaching the 100m terminal? The solution of one variable linear equation is used,

A and B practice the 100m race. They start from the starting point at the same time. A's speed is 7m / s, B's speed is 6.5m/s. After running for 10s, a sprained his foot, If the velocity is reduced to 6m / s, can b catch up with a before reaching the 100m terminal? The solution of one variable linear equation is used,

7*10=70
100-70=30
30/6=5
6.5*(10+5)=97.5

Party A and Party B practice the 100 meter race. They start from the starting point at the same time. The speed of Party A is 7 meters per second, and that of Party B is 6.5 meters per second. If Party A is allowed to run for one second first
A can catch up with B in a few seconds

"They ran for x seconds"
7x-6·5x=1
0·5x=1
0.5x divided by 0.5 = 1 divided by 0.5
x=2
2 times 7 = 14m 2 times 6.5 = 13m
14-13 = 1m

Party A and Party B practice the 100 meter race. They start from the starting point at the same time. The speed of Party A is 7 meters per second, and that of Party B is 6.5 meters
Party A and Party B practice the 100 meter race. They start from the starting point at the same time. The speed of Party A is 7 m / s, and that of Party B is 6.5 m / s. after running for 10 seconds, because party a twisted his foot, the speed was reduced to 6 m / S. can party B catch up with Party A before reaching the 100 meter finish? Don't copy

Before doing this topic, you should draw a line diagram on the draft paper. The line diagram is as follows:
At this time, a's foot twisted, and the speed dropped to 6m / s. B's speed did not change, so now it's B chasing A. their distance is 5m (see the picture), so now set: after X minutes, B will catch up with a
&Nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; &
&The solution is: x = 10
When x = 10, a has passed the 100 m terminal, and B has passed the 100 m terminal, so B can't catch up with a before reaching the 100 m terminal
If you don't understand,