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 with a velocity |t-4| where t is time in seconds the distance covered by the particle in 8 s is`
2 years ago

Sanju
106 Points
```							Since Im unable to upload the v vs t i will explain how the graph looks like. Velocity is taken on y-axix and time on x-axis. The graph will look like the letter M. It will form 2 right angled triangle. Triangle AOB & BCD. Here O is the origin and A is the point on y axis [positive axis]. OA = 4 m.. OB =4m..Coming to triangle BCD, CD is parallel to OA and is located at  a distance of 8 units from origin O in the x-axis..CD=4m and BD= 4m... Distance covered by the particle in 8 sec = area of triangle AOB + area of triangle BCD = 8+8 = 16 m
```
2 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