Home
Class 11
PHYSICS
A particle has initial velocity (2hati+3...

A particle has initial velocity `(2hati+3hatj)` and acceleration `(0.3hati+0.2hatj)`. The magnitude of velocity after 10 second will be

A

9 units

B

`9sqrt(2) units`

C

`5sqrt(2)units`

D

5 unit

Text Solution

Verified by Experts

The correct Answer is:
C
`vec(v)= vec(u)+vec(a)t`
`=(2hat(i)+3hat(j))+(0.3hat(i)+0.2hat(j))xx10`
`=(5hat(i)+5hat(j))implies |vec(v)|=5sqrt(2)units`
Promotional Banner

Topper's Solved these Questions

  • VECTORS

    A2Z|Exercise Chapter Test|29 Videos

  • VECTORS

    A2Z|Exercise Assertion Reasoning|17 Videos

  • UNIT, DIMENSION AND ERROR ANALYSIS

    A2Z|Exercise Chapter Test|28 Videos

  • WAVES AND ACOUSTICS

    A2Z|Exercise Chapter Test|30 Videos

Similar Questions

Explore conceptually related problems

A particle has an initial velocity of 3hati + 4hatj and on acceleration of 0.4hatj + 0.3hatj . The magnitude of its velocity after 10 s is

A particle has initial velocity (2 vec(i)+3 vec(j)) and acceleration (0.3 vec (i)+0.2 vec (j)) . The magnitude of velocity after 10 seconds will be

A particle has an initial velocity of 4 hati +3 hatj and an acceleration of 0.4 hati + 0.3 hatj . Its speed after 10s is

A particle has an initial velocity 3hati+4hatj and an accleration of 0.4hati+0.3hatj . Its speed after 10s is

A particle has an initial velocity of 3hati+ 4hatj and an acceleration of 0.4hati +0.3hatj Its speed after 10 s is

A particle has an initial velocity (6hati+8hatj) ms^(-1) and an acceleration of (0.8hati+0.6hatj)ms^(-2) . Its speed after 10s is

A particular has initial velocity , v=3hati+3hatj and a constant force F=4hati-3hatj acts on it. The path of the particle is

Given : vecA = 2hati + 3hatj and vecB = 5hati - 6hatj . The magnitude of vecA + vecB is

A body starts with a velocity (2hati+3hatj+11hatk)m//s and moves with an acceleration (5hati+5hatj-5hatk)m//s^(2) . What is its velocity after 0.2 sec ?