0

Guest

Courses New

please give me the answer to the third........its urgent

please give me the answer to the third........its urgent

Question Image
Grade:9

1 Answers

Arun
25750 Points
5 years ago
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) 
 = 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
Now substitute for t1 from (2) to get 
Distance covered = ½ *[a*b / (a +b)]*t²
 
Regards
Arun (askIITians forum expert)

Think You Can Provide A Better Answer ?

Other Related Questions on Mechanics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More