A car is moving on a straight road towards north with a uniform speed of 50 km/hour, then it turns left through 45° and it increases its speed at the same
time to 50/2 km/hour. Find the magnitude of change in the velocity of car.
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Answer:
Let the North be represented by Y-axis. The when the car turns left, it moves along the -ve x-axis, towards West.
Let \vec{v1} = 50 \hat{j}\ kmph\ \ \ and,\ \vec{v2} = -50 \hat{i}\ kmph
Change in velocity is = \vec{v2}-\vec{v1}=-50\hat{i}-50\hat{j}\ kmph
Magnitude of change in velocity is
50
2
+50
2
=50
2
The direction of the change in velocity = Southwest, 45 deg. from South towards West.
Explanation:
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