# Answer on Mechanics | Relativity Question for jojo

Question #4683

Tom and Jill are both robots that are going to have a race. Both start from rest. Tom has a constant acceleration for the entire run, while Jill does the following; first she accelerated for a time of 2.00 minutes(convert to seconds) at a rate of 2.20cm/s/s ( must convert to meters), then runs a distance of 18.9 kilometers reaching a final velocity of 23.7 m/s, thetn decelerates at a rate of 2.3m/s/s, finishing the race with a velocity of 6.00m/s.

If Tom ties with Jill, what would have to be his final velocity?

If Tom ties with Jill, what would have to be his final velocity?

Expert's answer

a- acceleration of Tom

a1=0.022m/sec^2-accel. Of Jill

1)velocity after two min of Jill =120*0.022=2.64m/sec

S=at^2/2=158m

2)distance of 18.9 kilometers reaching a final velocity of 23.7 m/s

s=& so a=2*18900/

at=23.7 m/s-2.64m/sec=21.06

s=& =t21.06/2

t=18900/10.53=1795sec

a=1.17m/sec^2 {at the distance of 18.9 kilometers reaching a final velocity of 23.7 m/s}

S=18900m

3) , thetn decelerates at a rate of 2.3m/s/s, finishing the race with a velocity of 6.00m/s

t*2.3=23.7-6

t=8sec.

S=vt-at^2/2=23.7*8-2.3*64/2=116m

total time- 120+1795+8sec=1923

total path = 158+18900+116=19174

4)Tom:

S=at^2/2

19174=a*1848964.5

a=0.01m/sec^2

final velocity=at=1923*0.01=19.23m/sec^2

a1=0.022m/sec^2-accel. Of Jill

1)velocity after two min of Jill =120*0.022=2.64m/sec

S=at^2/2=158m

2)distance of 18.9 kilometers reaching a final velocity of 23.7 m/s

s=& so a=2*18900/

at=23.7 m/s-2.64m/sec=21.06

s=& =t21.06/2

t=18900/10.53=1795sec

a=1.17m/sec^2 {at the distance of 18.9 kilometers reaching a final velocity of 23.7 m/s}

S=18900m

3) , thetn decelerates at a rate of 2.3m/s/s, finishing the race with a velocity of 6.00m/s

t*2.3=23.7-6

t=8sec.

S=vt-at^2/2=23.7*8-2.3*64/2=116m

total time- 120+1795+8sec=1923

total path = 158+18900+116=19174

4)Tom:

S=at^2/2

19174=a*1848964.5

a=0.01m/sec^2

final velocity=at=1923*0.01=19.23m/sec^2

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