Skip to content
Login Register

#vector acceleration

7 public questions tagged with this topic.

Motion Parameters and Calculations

A particle moves with a constant acceleration of (3i^−2j^)m/s2 from rest. What is the magnitude of its velocity after 4s

Velocity v=v0+at, with v0=0. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 14.42 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion Parameters and Calculations

View Details
Acceleration and Velocity Analysis

A particle moves in the x-y plane with a constant acceleration of (2 î + 3 ĵ) m/s². If it starts from rest at the origin

Velocity is given by v = v₀ + at, where v₀ = 0 (starts from rest). As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 7.21 m/s as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8

View Details
Acceleration and Velocity Analysis

A particle starts from rest with a constant acceleration of (4 î + 2 ĵ) m/s². What is its speed after 5 s?

Velocity v = v₀ + at, where v₀ = 0. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 10 m/s as the result, so option B is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Acceleration and Velocity Analysis - Part 8

View Details