MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 10
        
A car accelerates from rest at a constant rate 10 m/s2 for some time after which it decelerates at constant rate 5 m/s2 and then comes to rest. Total time elapsed is 15 s. Maximum velocity of the car is (i) while average velocity of the car during its deceleration is (ii) . Distance travelled by the car at the instant when its velocity is half of its maximum value is (iii) or (iv) .
10 months ago

Answers : (1)

Arun
22587 Points
							
 
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
10 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details