×

#### 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 train starts from rest and moves with a constant acceleration of 2m/s^2 for half a minute. The brakes are then applied and the train comes to rest in one minute after applying brakes. Find (a) the total distance moved by the train, (b) the maximum speed attained by the train and (c) the position of the train at the half the maximum speed.

```
2 years ago

Himanshu Rathour
119 Points
```							Let maximum speed attained by train is v, distance covered is s while accelerating and S while deaccelerating. v=0+2(30)=60m/ss=0+(1/2)(2)(900)s=900m0=60-a(60)a=1m/s²S=60(60)-½(60)²1S=3600-1800S=1800m
```
2 years ago
Himanshu Rathour
119 Points
```							Sorry I misunderstood the c part so here is the answer to (c) partDistance covered at half the maximum speed30²=0+2(2)s1s1=900/4=225m
```
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