Two Cars are in a Race. Car a passes the finish line 2 seconds before of Car B. Car A passes the finish line with a velocity V more than Car B. Both start the race with acceleration of 3m/s^2(Car A) and 2m/s^2(Car B).Calculate the value of V.
Divyankasc:
Yes..But did not have time to solve it..
I dont know it was right or wrong I did not have time to recheck...
you wasted your time -_-... don't attempt unless you are sure that you will be able to finish it
Lol...I thought it would be esy further...But it was not..
let me try
A very boring quadratic equation came in front..Now who will calculate the roots...
yep...
It would take time
It would have taken time...
yeah...
Answers
Answered by
2
Car A: u = 0, a = 3m/s², time = t, final velocity = v, distance =d
v = 0 + 3t = 3t
2ad =v² - u²
⇒ 2*3*d = (3t)²
⇒ 6d = 9t²
⇒ d = 1.5t²
Car B: u = 0, a = 2m/s², time = t+2, final velocity = v', distance = d
v' = 0 + 2(t+2) = 2t+4
2ad =v'² - u²
⇒ 2*2*d = (2t+4)²
⇒ 4d = 4t²+16t+16
⇒ d = t²+4t+4
So 1.5t² = t²+4t+4
⇒ 0.5t² - 4t - 4 = 0
⇒ t² - 8t - 8 = 0
⇒ t = 2(2+√6)s = 4+2√6 s
v = 3t = (12+6√6) m/s
v' = 2t+4 = 2(4+2√6)+4 = (12+4√6) m/s
V = v - v' = 2√6 m/s
v = 0 + 3t = 3t
2ad =v² - u²
⇒ 2*3*d = (3t)²
⇒ 6d = 9t²
⇒ d = 1.5t²
Car B: u = 0, a = 2m/s², time = t+2, final velocity = v', distance = d
v' = 0 + 2(t+2) = 2t+4
2ad =v'² - u²
⇒ 2*2*d = (2t+4)²
⇒ 4d = 4t²+16t+16
⇒ d = t²+4t+4
So 1.5t² = t²+4t+4
⇒ 0.5t² - 4t - 4 = 0
⇒ t² - 8t - 8 = 0
⇒ t = 2(2+√6)s = 4+2√6 s
v = 3t = (12+6√6) m/s
v' = 2t+4 = 2(4+2√6)+4 = (12+4√6) m/s
V = v - v' = 2√6 m/s
yep!
Very nice....thankuu
Was the other root negative?
you are most welcome!!:))
did you mark this option or any other?
I dont remember Now....
Q.P is not with me
no worries!! Hope it turns out correct!;)
there is a calculation mistake... please correct.
after correcting, i got 2 root 6
Answered by
1
s = u t + 1/2 a t²
first car: s = 0 + 1/2 * 3 * t²
2nd car: s = 0 + 1/2 * 2 * (t+2)²
From the two equations: t² - 8 t - 8 = 0
so t = 4 + 2 √6 sec
Now : v1 = 0 + 3 t
v2 = 0 + 2 (t+2)
subtract: V = t - 4 = 2 √6 m/sec
first car: s = 0 + 1/2 * 3 * t²
2nd car: s = 0 + 1/2 * 2 * (t+2)²
From the two equations: t² - 8 t - 8 = 0
so t = 4 + 2 √6 sec
Now : v1 = 0 + 3 t
v2 = 0 + 2 (t+2)
subtract: V = t - 4 = 2 √6 m/sec
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