Question

three sprinter A,B and C had to sprint from poiunts P to Q and back again ( starting in that order). The time interval between the starting times of the three sprinters A,B and C was 5 seconds each. Thus C started 10 seconds after A, while B started 5 seconds after A. The three sprinters passed a certain point R, which is somewhere between P and Q simultaneously. Having reached Q and reversed the direction, the third sprinter met the second one 9 m short of Q and met the first sprinter 15 m short of Q. find the speed of the first sprinter if the distance between PQ is equal to 55 m.

Answer 1

Answer:

Speed of the first sprinter A: 0.5 m/s

Step-by-step explanation:

Let a, b, and c be the speeds of the sprinters A, B, and C respectively.

When they passed a certain point R simultaneously, then

$\frac{PR}{a} -10=\frac{PR}{b} -5=\frac{PR}{c}\Rightarrow\frac{2}{b} =\frac{1}{a} +\frac{1}{c}$

Notice that b is the harmonic mean of a & c.

Similarly, we can write (using the above relation):

$\frac{55+9}{c} =\frac{55-9}{b}-5\Rightarrow\frac{64}{c} =\frac{46}{b}-5\Rightarrow\frac{23}{a} -\frac{41}{c} =5$

$\frac{55+15}{c} =\frac{55-15}{b}-10\Rightarrow\frac{70}{c} =\frac{40}{a}-10\Rightarrow\frac{4}{a} -\frac{7}{c} =1$

Solving the above simultaneous equations for a & c, we get

c=1 m/s & a=0.5 m/s

Also, b=2/3 m/s & PR=10 m

Hence, the speed of the first sprinter A is 0.5 m/s.

Answer 2

I appreciate your effort but the solution is incorrect .

the time taken by the three sprinters are different when they reach R .

here all the sprinters reach the point R simultaneously. That means although they reached the point R at the same point of time , their starting time were respectively different . There is a gap of 5 sec between all the 3 sprinters . So, if the distance between P and R is "y" , then all the three sprinters reach R by traversing "y" metres.

The main point of difference is their speed . "A" takes the highest time to reach R and hence it's speed will be the slowest , followed by "B" and finally "C" with the highest speed .

If we consider x m/s as the speed of "A" , then it would take "y/x" sec to reach R .

"B" takes "(y/x)-5" sec and "C" takes "(y/x)-10" sec to reach R .

From here we get the speed of B and C as (xy/y-5x) m/s and (xy/y-10x) m/s . (please use s=d/t)

The next part is child's play .

We just need to form 2 equations .

Now it says that when C reaches Q , it reverses the direction and starts moving towards P.

while moving , it meets B at a distance of 9m from Q.

the distance ratio would be 32:23 and it would be equal to the speed ratio of both these sprinters as the time is constant here .

a similar concept is used between C and A

the final equation would be ((y-5x)/(y-10x))=32/23

and ((xy/y-10x)/x)=7/4

i will not solve it as it is basic algebra . Anyone can do it .

the final answer is x = 1 m/s