Question

One runner in a relay race has a 1.50 s lead and is running at a constant speed of 3.25 m/s. The runner has 30.0 m to run before reaching the end of the track. A second runner moves in the same direction as the leader. What constant speed must the second runner maintain in order to catch up to the leader at the end of the race?

a. 3.9 m/s
c. 2.8 m/s
b. 4.75 m/s
d. 20 m/s

Answer 1

@dan815 @Deba_001 @Gravity_Dreams @eric_d @Elsa213 @RLeal @Luigi0210 @Future_UsArmy_MP @Thatkid_marc @iambatman @ikram002p

Answer 2

yes :)

Answer 3

=D

Answer 4

Can you help me out? /:
Kind of stuck..

Answer 5

@paki

Answer 6

@ParthKohli @pinal145 @Abhisar @AnswerMyQuestions @asatt32 @ams98 @davidza6496 @sammixboo @SarahEZZMcK @shkrina @SyedaLovesPie @sheelo_mughal @Stitch_ @zzr0ck3r @ziqbal103 @Zeta @zay813 @CO_oLBoY @CaseyCarns @chrisvillasmil @vksmith7 @victorbell @bloopman @beccamarx19 @bbbco123 @bbblakeee @ninjawolfjj @nuccioreggie @iambatman @OOOPS

Answer 7

I am waiting for you to work with.

Answer 8

okay... so would it be 4.75 m/s?

Answer 9

no, no, not that simple

Answer 10

be patient, please. Take steps and confirm if you got me

Answer 11

Alright.

Answer 12

the first one race at 3.25m/s so that if he is in front other in 1.5s , then the gap BA = 3.25*1.5 = 4.875 m, ok ?

Answer 13

so the total gap the second one have to run is 30+4.875=34.875 m

Answer 14

One more thing: from A , the time the first one have to spend to get the end is
t = \(\dfrac{30}{3.25}=9.23s\) ok?

Answer 15

Now, with that gap and that time, the second must run at v = \(\dfrac{34.875}{9.23}=3.78m/s\) And he has to catch up the first one before reaching the end, so, the speed should be 3.9m/s

Answer 16

Oooooh!! Got it! Thank you sooo much!!