A motor car is going due north at a speed of $50$ km/hr. It makes a $90^{o}$ left turn without changing the speed. The change in the velocity of the car is about.

50 \sqrt{2}

km/hr

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70

km/hr

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80

km/hr

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Zero

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Solution

The correct option is B $50 \sqrt{2}$ km/hr
change in velocity = final velocity - initial velocity
initial velocity = 50 kmph north
final velocity =50 kmph west
therefore change in velocity = $\sqrt{50^{2} + 50^{2}} = 50 \sqrt{2} k m p h$

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Similar questions

Q. A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. The change in the velocity of the car is about
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(c) 70 km/h towards north-west
(d) zero.

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A motor car is going due north at a speed of 50km/hr. It makes a 90o left turn without changing the speed. The change in the velocity of the car is about.

The correct option is B 50√2km/hrchange in velocity = final velocity - initial velocityinitial velocity = 50 kmph northfinal velocity =50 kmph west therefore change in velocity =√502+502=50√2kmph

A motor car is going due north at a speed of $50$ km/hr. It makes a $90^{o}$ left turn without changing the speed. The change in the velocity of the car is about.

The correct option is B $50 \sqrt{2}$ km/hr
change in velocity = final velocity - initial velocity
initial velocity = 50 kmph north
final velocity =50 kmph west
therefore change in velocity = $\sqrt{50^{2} + 50^{2}} = 50 \sqrt{2} k m p h$