×

#### Thank you for registering.

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

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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
```
A particle moves in a straight line and its position x at the time t is given by x^=2+t its acceleration is given by
A particle moves in a straight line and its position x at the time t is given by x^=2+t its acceleration is given by

```
3 years ago

Arun
25768 Points
```							X^2 = 2 + td(x^2) / dv = d(2 + t)/dv2dx/dv = dt/dv2xdx = dtdx/dt = 1/(2x)v = 1/(2x)Differentiating above equationdv/dx = -1/(2x^2)Acceleration is given bya = vdv/dx= 1/(2x) * -1/(2x^2)= -1/(4x^3)It’s acceleration is given by -1/(4x^3)
```
3 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »

### Course Features

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

### Course Features

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