badge image

Enroll For Free Now & Improve Your Performance.

×
User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
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: 12th pass

                        

The relation between acceleration and time for the object is given by a=2t+4t^2. Calculate the position of the object at t=4sec?

3 years ago

Answers : (1)

Arun
24742 Points
							
We need the position and velocity at t=0. Call these x₀ and v₀. They may be zero in your question but you haven't stated this. 

a = 2t + 4t² 

Since a = dv/dt: 
dv/dt = 2t + 4t² 

Integrate: 
v = t² + (4/3)t³ + c 

When t = 0, v = v₀ so: 
v₀ = 0² + (4/3)0³ + c 
c = v₀ 

v = t² + (4/3)t³ + v₀ 

Since v = dx/dt: 
dx/dt = t² + (4/3)t³ + v₀ 

Integrate: 
x = (1/3)t³ + (4/3)(1/4)t⁴ + v₀t + C 
. .= (1/3)t³ + (1/3)t⁴ + v₀t + C 

When t = 0, x = x₀ so: 
x₀ = (1/3)0³ + (1/3)0⁴ + v₀0 + C 
x₀ = C 

x = (1/3)t³ + (1/3)t⁴ + v₀t + x₀ 

When t=4s: 
x = (1/3)4³ + (1/3)4⁴ + 4v₀ + x₀ 
. = 64/3 + 256/3 + 4v₀ + x₀ 

If we take x₀ = 0 and v₀=0 this gives: 
x = (64 + 256)/3 
x = 320/3 metres
3 years 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

  • 18 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