22 public questions tagged with this topic.
Velocity v=v₀+at. Starting from rest, v=at. With a=(ax i+ay j), magnitude |v|=t√(ax²+ay²) per NCERT vector addition. Substituting ax, ay, t gives |v| equal to It remains constant. Hence option A is correct.
Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Motion in a Plane and Projectile Motion
Displacement r = v₀ t + (1/2) a t². 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 9 m 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
Velocity components: vx=5m/s,vy=12m/s. 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 13 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
Displacement r=v0t+12at2. 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 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
Velocity v=v0+at. 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 6 m/s as the result, so option C is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Mixed Concepts
Velocity v=v0+at. 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 13 m/s as the result, so option D is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Mixed Concepts
Velocity: vx=dxdt=2+2t,vy=dydt=3−4t. 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 6 m/s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving
Velocity: vx=dxdt=6t−4,vy=dydt=5. 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 4 m/s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving
For free fall, s=½gt² and v²=2gh per NCERT. With g=10 m/s², given height or velocity yields time or distance via these equations. Substituting values gives result matching 15 m/s. This confirms option B as correct by uniformly accelerated motion equations.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving
Use v=v0+at. Here, v0=26m/s, a=−10m/s2, t=2s. 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 6 m/s as the result, so option B is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Uniform Acceleration and Graphical Analysis
Use the equation v=v0+at. 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 12 m/s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion in a Straight Line - Advanced Problems
Use the equation v=v0+at. 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 21 m/s as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Free Fall and Vertical Motion
