please give me the answer to the third........its urgent
please give me the answer to the third........its urgent
gurnoor kaur , 7 Years ago
Grade 9
Mechanics> please give me the answer to the third......
1 Answers
Arun
Dear student
I have taken a and b in place of alpha and beta.
Let the car accelerate for t1sec at a m/sec² and decelerate for t2sec. at - b/ sec²
At the end of t1 sec. its velocity = 0 + at1 m/sec. At the end of t2 sec its velocity = 0
Initial velocity at the beginning of deceleration is α t1
V(t2) = a t1 - b t2 = 0 Therefore t2 = (a/ b ) t1 ------------------------(1)
But t = t1 + t2 ⇒ t = t1 + t2 = t1 + (a/ b ) t1 = t1(1+ a/ b)
∴ t1 = t / (1+ a/ b) = {b / (a + b)}*t -------------------------------------- (2)
Max. Velocity reached = a t1 = a t / (1+ a/ b) = t / (1/ a + 1 / b)
At the end of t1 sec. its velocity = 0 + at1 m/sec. At the end of t2 sec its velocity = 0
Initial velocity at the beginning of deceleration is α t1
V(t2) = a t1 - b t2 = 0 Therefore t2 = (a/ b ) t1 ------------------------(1)
But t = t1 + t2 ⇒ t = t1 + t2 = t1 + (a/ b ) t1 = t1(1+ a/ b)
∴ t1 = t / (1+ a/ b) = {b / (a + b)}*t -------------------------------------- (2)
Max. Velocity reached = a t1 = a t / (1+ a/ b) = t / (1/ a + 1 / b)
= a*b*t/(a+b)
Distance covered =(a t1² + b t2²) /2 = ={a t1² + b [(a/ b ) t1)²]} /2 (substitute for t2 from (1)
Simplifying this = ½ * (a/ b) * (a + b)* t1²
Distance covered =(a t1² + b t2²) /2 = ={a t1² + b [(a/ b ) t1)²]} /2 (substitute for t2 from (1)
Simplifying this = ½ * (a/ b) * (a + b)* t1²
Now substitute for t1 from (2) to get
Distance covered = ½ *[a*b / (a +b)]*t²
Distance covered = ½ *[a*b / (a +b)]*t²
Regards
Arun (askIITians forum expert)
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