Question

V4 MULTIPLE CHOICE [ +4.0/-1.0]
Attempted Questions:2/54
Two cars, A and B, start from rest accelerating at 2m/s² and 6m/s- respectively. After covering 100 m, each
car immediately starts decelerating at 4 m/s until it comes to rest. Choose the correct choice(s).
DETE
Options
01 ) car A covers a total distance 150m
02 ) car B covers a total distance 250 m
03 ) car A moves for total time 15 seconds
04 ) car B moves for total time 25 seconds
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Answer 1

Answer:

01 ) car A covers a total distance 150m

02 ) car B covers a total distance 250 m

03 ) car A moves for total time 15 seconds

Explanation:

1) For A:

From A to A1,

Using Newton's Equation of motion,

$v^2-u^2=2as\\ \\=>v^2-0^2=2*2*100\\ \\=>v=20m/s$

And,

$v=u+at\\ \\=>20=0+2*t\\ \\=>t_1=10s$

Also,

From A1 to A2,

$v^2-u^2=2as\\ \\=>0^2-20^2=2*-4*s\\ \\=>s=50m$

And,

$v=u+at\\ \\=>0=20-4t\\ \\=>t=5s$

Hence, total time taken by A is  10+5=15s.

Total distance covered by A is 100+50=150m .

=> Option 1) ,3) are correct.

2) For B:

From B to B1,

Using Newton's equation of motion,

$v^2-u^2=2as\\ \\=>v^2-0^2=2*6*100\\ \\=>v=20\sqrt{3}m/s$

And,

$v=u+at\\ \\=>20\sqrt{3}=0+6*t\\ \\=>t_1=\frac{10}{\sqrt{3}}s$

Also,

From A1 to A2,

$v^2-u^2=2as\\ \\=>0^2-(20\sqrt{3})^2=2*-4*s\\ \\=>s=150m$

And,

$v=u+at\\ \\=>0=20\sqrt{3}-4t\\ \\=>t=5\sqrt{3}s$

Hence, total time taken by B is

$5\sqrt{3}+ \frac{10}{\sqrt{3}}=\frac{25}{\sqrt{3}}$

Total distance covered by B is 100+150=250m .

Hence, Finally option (1) , (2) and (3) are correct.