A bus is moving on a straight road towards north with a uniform speed ...
The speed is constant .i.e. acceleration =0.
Now initial velocity = u = +50j
and final velocity = v = -50i
Change in velocity ===>
v – u = -50i – 50 j = 50(-i – j) km/hr.
magnitude = 50*root2. = 70.7 km/hr.
Direction u can check as it comes:
-i-j
that is Sout west.
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A bus is moving on a straight road towards north with a uniform speed ...
Analysis:
To find the increase in velocity of the bus during the turning process, we need to consider the change in direction of the bus. Since the speed of the bus remains unchanged, the increase in velocity is solely due to the change in direction.
Initial velocity:
Magnitude: 50 km/hr
Direction: North
Change in direction:
The bus turns through 90 degrees, which means its new direction is perpendicular to its initial direction.
Final velocity:
Magnitude: 50 km/hr (unchanged)
Direction: East (perpendicular to the initial direction)
Calculating the increase in velocity:
To calculate the increase in velocity, we need to find the vector difference between the initial and final velocities.
Step 1: Converting velocities to vector form:
Initial velocity vector: 50 km/hr * (0, 1) (since it is moving towards the north)
Final velocity vector: 50 km/hr * (1, 0) (since it is now moving towards the east)
Step 2: Finding the difference between the vectors:
Final velocity vector - Initial velocity vector = (1, 0) - (0, 1) = (1-0, 0-1) = (1, -1)
Step 3: Calculating the magnitude and direction:
Magnitude of the increase in velocity = sqrt(1^2 + (-1)^2) = sqrt(2) ≈ 1.41 km/hr
Direction of the increase in velocity = arctan(-1/1) = -45 degrees (measured counterclockwise from the positive x-axis)
Therefore, the increase in velocity of the bus during the turning process is approximately 1.41 km/hr in a direction of -45 degrees (or southwest).
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