10 public questions tagged with this topic.
Speed: 90km/h=25m/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 730 m 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
Convert speed: 18km/h=18⋅10003600=5m/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 5 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
Phase 1: v=4⋅6=24m/s, x1=12⋅4⋅(6)2=72m. 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 192 m as the result, so option D is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Equations of Motion and Problem Solving
Use x=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 15 m as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Uniform Acceleration and Graphical Analysis
Use y=12gt2. Here, g=10m/s2, t=3s. 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 45 m as the result, so option D is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Uniform Acceleration and Graphical Analysis
For constant speed, distance x=vt. 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 300 m as the result, so option D is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion in a Straight Line - Advanced Problems
For constant speed, distance x=vt. 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 360 m as the result, so option D is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations
Find a=v−v0t=0−279=−3m/s2. 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 100 m as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations
Phase 1: v=10+1⋅5=15m/s, x1=10⋅5+12⋅1⋅(5)2=50+12.5=62.5m. 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 115 m as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations
Phase 1: v=5⋅4=20m/s, x1=12⋅5⋅(4)2=40m. 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 95 m as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Distance, Displacement and Velocity-Time Relations
