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
Hrishant Goswami Grade: 10
        The acceleration of a cart started at t = 0, varies with time as shown in figure (3-E2). Find the distance travelled in 30 seconds and draw the position-time graph.
3 years ago

Answers : (1)

Jitender Pal
askIITians Faculty
365 Points
										Sol. In 1st 10 sec S1 = ut + 1/2 at2 ⇒ 0 + (1/2 x 5 x 102 ) = 250 ft.

At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec.
∴ From 10 to 20 sec (∆t = 20 – 10 = 10 sec) it moves with uniform velocity 50 ft/sec,
Distance S2 = 50 × 10 = 500 ft
Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s2. At 20 sec velocity is 50 ft/sec.
t = 30 – 20 = 10 s
S3 = ut + 1/2 at2
= 50 x 10 + (1/2) (-5) (10)2 = 250 m
Total distance travelled is 30 sec = S1 + S2 + S3 = 250 + 500 + 250 = 1000 ft.
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
Get extra Rs. 413 off
USE CODE: COUPON10
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details
Get extra Rs. 445 off
USE CODE: COUPON10

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details