A body travels \( 102.5 \, \text{m} \) in \( n^\text{th} \) second and \( 115.0 \, \text{m} \) in \( (n+2)^\text{th} \) second. The acceleration is:
- \( 9 \, \text{m/s}^2 \)
- \( 6.25 \, \text{m/s}^2 \)
- \( 12.5 \, \text{m/s}^2 \)
- \( 5 \, \text{m/s}^2 \)
The Correct Option is B
Approach Solution - 1
Formula for Distance Travelled in the nth Second
The distance sn travelled by a body in the nth second is given by:
sn = u + (\(\frac{a}{2}\))(2n − 1)
where u is the initial velocity, a is the acceleration, and n is the specific second.
Given Data:
Distance travelled in the nth second: sn = 102.5 m
Distance travelled in the (n + 2)th second: sn+2 = 115.0 m
Using the Formula for the (n + 2)th Second:
sn+2 = u + $\frac{a}{2}$(2n + 3)
Set Up Equations for sn and sn+2:
From the given distances:
102.5 = u + $\frac{a}{2}$(2n − 1)
115.0 = u + $\frac{a}{2}$(2n + 3)
Subtract the First Equation from the Second:
115.0 − 102.5 = (u + $\frac{a}{2}$(2n + 3)) − (u + $\frac{a}{2}$(2n − 1))
12.5 = $\frac{a}{2}$ × 4
12.5 = 2a
a = $\frac{12.5}{2}$ = 6.25 m/s²
Conclusion:
The acceleration of the body is 6.25 m/s².
Approach Solution -2
The distance travelled in the \( n^{th} \) second, \( s_n = 102.5 \, \text{m} \)
The distance travelled in the \( (n+2)^{th} \) second, \( s_{n+2} = 115.0 \, \text{m} \).
We are asked to find the acceleration \( a \).
Step 2: Formula for distance travelled in the \( n^{th} \) second.
The distance travelled in the \( n^{th} \) second under uniform acceleration is given by: \[ s_n = u + \frac{a}{2}(2n - 1) \] where \( u \) is the initial velocity and \( a \) is the acceleration.
Step 3: Write equations for both cases.
For the \( n^{th} \) second: \[ s_n = u + \frac{a}{2}(2n - 1) = 102.5 \] For the \( (n + 2)^{th} \) second: \[ s_{n+2} = u + \frac{a}{2}[2(n + 2) - 1] = u + \frac{a}{2}(2n + 3) = 115.0 \]
Step 4: Subtract the two equations.
\[ s_{n+2} - s_n = \frac{a}{2}[(2n + 3) - (2n - 1)] = \frac{a}{2}(4) = 2a \] \[ 115.0 - 102.5 = 12.5 = 2a \] \[ \Rightarrow a = 6.25 \, \text{m/s}^2 \]
Step 5: Final Answer.
\[ \boxed{a = 6.25 \, \text{m/s}^2} \]
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 Kinematics Questions
- A body falling under gravity covers two points A and B separated by 80 m in 2s. The distance of upper point A from the starting point is ________ m (use g = 10 ms–2)
- Position of an ant \( S \) (in meters) moving in the Y-Z plane is given by \( S = 2t \hat{j} + 5t \hat{k} \) (where \( t \) is in seconds). The magnitude and direction of velocity of the ant at \( t = 1 \, s \) will be:
- A particle starts from origin at \( t = 0 \) with a velocity \( 5\hat{i} \, \text{m/s} \) and moves in the \( x\)-\( y \) plane under the action of a force that produces a constant acceleration of \( (3\hat{i} + 2\hat{j}) \, \text{m/s}^2 \). If the \( x \)-coordinate of the particle at that instant is 84 m, then the speed of the particle at this time is \( \sqrt{\alpha} \, \text{m/s} \). The value of \( \alpha \) is ______.
- A particle is moving in one dimension (along x axis) under the action of a variable force. It's initial position was 16 m right of origin. The variation of its position (x) with time (t) is given as x = –3t3 + 18t2 + 16t, where x is in m and t is in s. The velocity of the particle when its acceleration becomes zero is _________ m/s.
- The relation between time \( t \) and distance \( x \) is \( t = \alpha x^2 + \beta x \), where \( \alpha \) and \( \beta \) are constants. The relation between acceleration \( a \) and velocity \( v \) is:
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 ______.