QR question

[299678]

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Q. A foot baller tries to reach a ball moving at 3 m/s, if he is running in the same direction 5 seconds after the ball crossed him, at what speed should he ran to reach the ball within 5 seconds?

Could you be able to solve this problem by setting up this?

d r t
ball
player
total

thank you

 

[sfoksn]

Q. A foot baller tries to reach a ball moving at 3 m/s, if he is running in the same direction 5 seconds after the ball crossed him, at what speed should he ran to reach the ball within 5 seconds?

Could you be able to solve this problem by setting up this?

d r t
ball
player
total

thank you

Well, since after 5 seconds, the football, moving 3 m/s per second, it wouuld have passed him 15 meters. So the player has to catch up 15 meters total in 5 seconds. So that means he needs to get ahead at 5 m/s, which means he needs to run at 8m/s.

I think..

 

[RCT PC CRN]

Ball travels 30 m in total at 3 m/s (15 m before foot baller start running and 15 after foot baller start running). Foot baller has to compete same distance in 5 sec which bring his speed to 6 m/s.

 

[RCT PC CRN]

Sorry I forgot to give you in form of d r t.

For ball -

d = 3 m/s * 10 sec (5 before and 5 after) = 30 m

For Foot baller

d = same as ball = 30 m = r * 5 sec (Remember foot baller has only 5 sec)

r = 6 m/s

 

[sfoksn]

what is the answer, and i cannot visualize the problem. am i wrong about the situation described in the problem?

 

[Lethstang]

ok

the balls moving at 3 m/s

you dont start running after it till five seconds pass. so in that five seconds, the ball is 15m away.

you now have five seconds to reach it. in that five seconds, the ball moves another 15m, bringing total distance traveled to 30 meters

so you now have to traverse 30m in 5s to catch up to the ball. thats 30/5, so 6 m/s.

 

[sfoksn]

haha.. i just looked back at my answer and yes i messed up simple division. i thought 15/5 is 5..