Question #206106

Balls A and B accelerate from rest towards each other at 2 m/s2 and 3 m/s2 respectively. Given that the initial distance between the balls is 1500 m, determine:


i. the time taken for the balls to meet.

ii. the distance covered by ball A when the balls meet.

iii. the distance travelled by ball B when the balls meet.


1
Expert's answer
2021-06-14T14:49:55-0400

Gives

a1=2m/seca_1=2m/sec

a2=3m/sec2a_2=3m/sec^2

Distance (s)=1500m1500m

Ball A

Time take t1

s=ut+12at2s=ut+\frac{1}{2}at^2

s=1500-x

u=0u=0

t1=2sa1t_1=\sqrt\frac{2s}{a_1}

t1=2(1500x)2t_1=\sqrt\frac{2(1500-x)}{2}

Second ball

s=x

t2=2×xa2t_2=\sqrt\frac{2\times x}{a_2}

t1=t2t_1=t_2

2(1500x)2=2(x)3\sqrt\frac{2(1500-x)}{2}=\sqrt\frac{2(x)}{3}

Both side squre take

1500x=2x31500-x=\frac{2x}{3}

45003x=2x4500-3x=2x

5x=45005x=4500

x=900mx=900m

Both ball are meet same time

t1=t2=t=2×9003=600sect_1=t_2=t=\sqrt\frac{2\times900}{3}=\sqrt{600}sec


Part (b)

Distance covered by ball A

x=1500900=600mx'=1500-900=600m

Part( c)

Dixtance covered ball B

x=x=900mx=x''=900m



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS