Problem 4 (1 point) Two cyclists A and B are racing two races on the straight tr

Question

Problem 4 (1 point) Two cyclists A and B are racing two races on the straight track. The cyclists are starting from the same initial point x-0. Race 1: Cyclist A is initially at rest, and starts accelerating with a constant acceleration aa. Cyclist B starts moving at a constant velocity vs. Cyclist A catches up with cyclist B at xi 10 m Race 2: Cyclist A is initially at rest, and starts accelerating with the same constant acceleration a' aA Cyclist B starts moving with the constant velocity equal to twice that from the first race v 2 Va. Cyclist A catches up with cyclist B at position x2. Coordinate x2 is equal to: (E) x2 5 m (F) x 2.5 m (A) x2 40 m (B) x2 20 m (C) x- 10 V2 m (D) x2 10 m (H) None of the above.A

Explanation / Answer

For race 1:

for Cyclist 1, distance traveled is 10 m when cyclist two catches him, So

Using equation

d = U*t + 0.5*a*t^2

10 = 0*t + 0.5*aA*t^2

aA = 20/t^2

for cyclist 2, distance traveled in the same time period will be

distance = speed*time

10 = Vb*t

Vb = 10/t

Now from above equation

t = 10/Vb

aA = 20/t^2 = 20/(10/Vb)^2

aA = Vb^2/5

Now For race 2:

Given that cyclist A catches up with B at x2, So again suppose time taken to catch is t1, then distance traveled will be

for cyclist A

d = U*t + 0.5*a*t^2

x2 = 0*t1 + 0.5*aA*t1^2

for cyclist B:

d = V*t

x2 = 2*Vb*t1

Since both have traveled same distance, So

0.5*aA*t1^2 = 2*Vb*t1

Now we know that

aA = Vb^2/5, So

(1/2)*[(Vb^2)/5]*t1^2 = 2*Vb*t1

Vb*t1 = 20

Using this

x2 = 2*Vb*t1

x2 = 2*20 = 40 m

Correct option is A.

Please Upvote. Comment below if you have any doubt.

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