The area under the velocity-time graph represents the displacement of an object. It is calculated by integrating the velocity with respect to time or finding the area geometrically in cases of simple shapes. For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/cRead more
The area under the velocity-time graph represents the displacement of an object. It is calculated by integrating the velocity with respect to time or finding the area geometrically in cases of simple shapes.
Using s = ut +1/2at², where u = 0, a = 4m/s² and t = 5s: s = 0 +1/2(4)(5²) = 1/2(4)(25) = 50 × 2 = 100m. This question related to Chapter 2 physics Class 11th NCERT. From the Chapter 2. Motion in Straight Line. Give answer according to your understanding. For more please visit here: https://www.tiwaRead more
Using s = ut +1/2at², where u = 0, a = 4m/s² and t = 5s:
s = 0 +1/2(4)(5²) = 1/2(4)(25) = 50 × 2 = 100m.
This question related to Chapter 2 physics Class 11th NCERT. From the Chapter 2. Motion in Straight Line. Give answer according to your understanding.
The second equation of motion is s=ut+ 1/2at², where u is the initial velocity, a is the acceleration, and t is the time. For objects starting from rest (u =0), the displacement simplifies to s = 1/2at². For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chaptRead more
The second equation of motion is s=ut+ 1/2at², where u is the initial velocity, a is the acceleration, and t is the time. For objects starting from rest (u =0), the displacement simplifies to s = 1/2at².
The second equation of motion is s = ut +1/2at². Substituting u = 0, s = 20m and t = 2s: 20 = 0 +1/2a(2²) 20 = 2a a = 10 m/s². For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
The second equation of motion is s = ut +1/2at². Substituting u = 0, s = 20m and t = 2s:
20 = 0 +1/2a(2²)
20 = 2a
a = 10 m/s².
The first equation of motion is v = u + at. Here, u = 0, a = 5m/s² and t = 4s. Substituting these values: v = 0 + 5 × 4 = 20 m/s. For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
The first equation of motion is v = u + at. Here, u = 0, a = 5m/s² and t = 4s. Substituting these values:
v = 0 + 5 × 4 = 20 m/s.
What does the area under the velocity-time graph of an object represent?
The area under the velocity-time graph represents the displacement of an object. It is calculated by integrating the velocity with respect to time or finding the area geometrically in cases of simple shapes. For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/cRead more
The area under the velocity-time graph represents the displacement of an object. It is calculated by integrating the velocity with respect to time or finding the area geometrically in cases of simple shapes.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
An object moves with uniform acceleration of 4 m/s². If it starts from rest, what will be its displacement after 5 seconds?
Using s = ut +1/2at², where u = 0, a = 4m/s² and t = 5s: s = 0 +1/2(4)(5²) = 1/2(4)(25) = 50 × 2 = 100m. This question related to Chapter 2 physics Class 11th NCERT. From the Chapter 2. Motion in Straight Line. Give answer according to your understanding. For more please visit here: https://www.tiwaRead more
Using s = ut +1/2at², where u = 0, a = 4m/s² and t = 5s:
s = 0 +1/2(4)(5²) = 1/2(4)(25) = 50 × 2 = 100m.
This question related to Chapter 2 physics Class 11th NCERT. From the Chapter 2. Motion in Straight Line. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
If an object starts from rest and moves with uniform acceleration, what will be its displacement after 𝑡 seconds?
The second equation of motion is s=ut+ 1/2at², where u is the initial velocity, a is the acceleration, and t is the time. For objects starting from rest (u =0), the displacement simplifies to s = 1/2at². For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chaptRead more
The second equation of motion is s=ut+ 1/2at², where u is the initial velocity, a is the acceleration, and t is the time. For objects starting from rest (u =0), the displacement simplifies to s = 1/2at².
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
A particle covers 20m in the first 2 seconds of its motion from rest under uniform acceleration. What is the magnitude of its acceleration?
The second equation of motion is s = ut +1/2at². Substituting u = 0, s = 20m and t = 2s: 20 = 0 +1/2a(2²) 20 = 2a a = 10 m/s². For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
The second equation of motion is s = ut +1/2at². Substituting u = 0, s = 20m and t = 2s:
20 = 0 +1/2a(2²)
20 = 2a
a = 10 m/s².
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
A car starts from rest and accelerates uniformly at 5m/s². What will be its velocity after 4 seconds?
The first equation of motion is v = u + at. Here, u = 0, a = 5m/s² and t = 4s. Substituting these values: v = 0 + 5 × 4 = 20 m/s. For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
The first equation of motion is v = u + at. Here, u = 0, a = 5m/s² and t = 4s. Substituting these values:
v = 0 + 5 × 4 = 20 m/s.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/