Question

Two soccer players start from rest, 36 m apart. They run directly toward each other, both players accelerating. The first player’s acceleration has a magnitude of 0.55 m/s2. The second player’s acceleration has a magnitude of 0.41 m/s2. (a) How much time passes before the players collide? (b) At the instant they collide, how far has the first player run?

Answer 1

a.) S1 = 1/2 × a1 t^2 = 0.55/2 ×t^2 = 0.275t^2

Similarly , S2 = 0.41/2 ×t^2 = 0.205 t^2

Also according to question ,

S1 + S2 = 36

=) 0.48 t^2 = 36

=) t = √ (36/0.48) = 8.66s

b.) S1 = 0.275 t^2 = 0.275×(8.66)^2 = 20.625m

That means distance travel by first player = 20.6m

Answer 2

SOLUTION :

a.

Let they collide after t secs. 

So,

As they both start from the rest :

Distance travelled by first player = 1/2 a t^2 = 1/2 * 0.55 * t^2 = 0.275 t^2

Distance travelled by second payer = 1/2 a’ t^2 = 1/2 * 0.41* t^2 = 0.205^2

As they collide, the sum of their distances travelled = 36 m

=> 0.275 t^2 + 0.205 t^2 = 36

=> 0.48 t^2 = 36

=> t^2 = 36 / 0.48 = 75 

=> t = 8.66 secs.

Players collide after 8.66 secs (ANSWER).

b.

First player’s distance travelled = 0.275 * 75= 20.625 m .

Hence,  they collide at a  point 20.625 m  from the starting point of the first player. (ANSWER)