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 starting from rest moves in a straight line with acceleration as shown in the a-t graph.Find the distance(in m) traveled by the particle in the first four seconds from start of its motion.

```
5 years ago

```							first make v-t graph then calculate area under that graph
```
5 years ago
```							When t=0,then a=1When a=1,then t=0 When time=4,thena=4Given u=0S=ut+1/2at^2S=1/2(4)(4)^2Then,S=32
```
one year ago
```							Here we have to put this graph in the equation of straight line as Y,=MX+c.with negative slope.here we will get a(t)=1-t.now we integrate it to find velocity which will be v=t-t2/2(here t2 means t square).now again we will integrate it to get distance covered which will be s=t2/2-t3/6.now we will put t=1 ...2...3....and 4 to get distance covered and then we will add all of them to get total distance covered..answer would be 4.01m.
```
one year 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