A train starting from rest first accelerates uniformly up to a speed of 80 km/h for time t, then it moves with a constant speed for time 3t. The average speed of the train for this duration of journey will be (in km/h) :
- 80
- 70
- 30
- 40
The Correct Option is B
Approach Solution - 1
The average speed is calculated using the formula:
\[ \text{Average speed} = \frac{\text{total distance}}{\text{time taken}} \]
During the first phase of acceleration:
\[ \text{Distance covered} = \frac{1}{2} \times \text{final speed} \times \text{time} = \frac{1}{2} \times 80 \times t = 40t \]
During the second phase of constant speed:
\[ \text{Distance covered} = \text{speed} \times \text{time} = 80 \times 3t = 240t \]
Total distance covered:
\[ 40t + 240t = 280t \]
Total time taken:
\[ t + 3t = 4t \]
Average speed:
\[ \text{Average speed} = \frac{280t}{4t} = 70 \, \text{km/h} \]
Approach Solution -2
To calculate the average speed of the train for the entire journey, we need to consider the two phases of motion: acceleration and constant speed.
- The train accelerates uniformly to a speed of 80 km/h. Let's assume the acceleration occurs over a time \( t \). During this phase, the average speed is half of the final speed due to uniform acceleration from rest. Hence, the average speed during acceleration is: \(\frac{80}{2} = 40 \text{ km/h}\).
- Next, the train moves at a constant speed of 80 km/h for a time duration of \( 3t \).
The average speed for the entire journey is calculated by dividing the total distance traveled by the total time taken.
Firstly, calculate the distance covered in each phase:
- Distance during acceleration: Since the speed increases uniformly from rest to 80 km/h over time \( t \), the distance covered is: \(d_1 = \text{average speed} \times \text{time} = 40 \times t = 40t \text{ km}\).
- Distance at constant speed: The train travels at 80 km/h for \( 3t \) hours, so the distance covered during this period is: \(d_2 = 80 \times 3t = 240t \text{ km}\).
The total distance \( D \) is: \(D = d_1 + d_2 = 40t + 240t = 280t \text{ km}\).
The total time \( T \) taken for the journey is the sum of the two time intervals: \(T = t + 3t = 4t \text{ hours}\).
Thus, the average speed \( V \) for the entire journey is given by: \(V = \frac{D}{T} = \frac{280t}{4t} = 70 \text{ km/h}\).
Therefore, the average speed of the train over the duration of the journey is 70 km/h.
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Acceleration Questions
- solid sphere is rolling purely with speed v on horizontal surface. It rolls up an incline surface and stops at height h. Then height h is [g is the acceleration due to gravity]:
- In the system shown below, the pulley \(4\) string are ideal. If the acceleration of blocks is \(\frac{g}{8}\) , find \(\frac{m_1}{m_2}\)
- The acceleration due to gravity on the surface of earth is g. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be:
- Find the acceleration of \(2\) \(kg\) block shown in the diagram (neglect friction)
- A particle is having uniform acceleration. If its displacement from t to \((t + 1)\) second is \(120 \) \(m\) and change in velocity is \(50 \;m/s\). Find its displacement (in m) in \((t + 2)^{th}\) second
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.