Force Acting on the Block:
The force \( F \) acting on the block increases linearly with time \( t \).
We can write:
\[ F = kt \] where \( k \) is a constant of proportionality.
Using Newton's Second Law to Find Acceleration:
According to Newton's second law, the acceleration \( a \) of the block is given by:
\[ F = ma \]
Substituting \( F = kt \):
\[ ma = kt \implies a = \frac{kt}{m} \]
where \( m \) is the mass of the block.
Relation Between Acceleration and Time:
From the equation \( a = \frac{kt}{m} \), we see that the acceleration \( a \) is directly proportional to time \( t \).
This means that as time \( t \) increases, the acceleration \( a \) also increases linearly.
Conclusion:
The correct graph representing the acceleration of the block with time is a straight line passing through the origin, as shown in Option (2).
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).