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:
\[ \text{Average speed} = \frac{280t}{4t} = 70 \, \text{km/h} \]
The acceleration due to gravity at a height of 6400 km from the surface of the earth is \(2.5 \, \text{ms}^{-2}\). The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth = 6400 km)
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: