Question

A particle travels with uniform velocity of 5m/s, due north for 3 seconds. It stops in negligible time and then travels with uniform velocity of 10 m/s, due south, for 7 seconds. The average velocity of the particle is (only magnitude)​

Answer 1

Answer:

5.5 m/s

Step-by-step explanation:

distance1= 5x3=15

distance2=(-10)x7=(-70)

avg velocity= total distance/total time

Answer 2

The average velocity of the particle is 5.5 m/s

Step-by-step explanation:

When the particle travels due North with velocity of 5 m/s for 3 seconds

Then, the displacement of particle in North direction

$=3\times 5$

$=15$ m

Similarly,

When the particle travels due South with velocity of 10 m/s for 7 seconds

Then, the displacement of particle in South direction

$=10\times 7$

$=70$ m

If we take displacement in South direction as positive then the displacement in North direction will be negative

Therefore,

Total displacement = $70-15=55$ m

Total time = $3+7=10$ seconds

Therefore, the average velocity

$\text{Average Velocity}=\frac{\text{Total Displacement}}{\text{Total Time}}$

$\implies \text{Average Velocity}=\frac{55}{10}$

$\implies \text{Average Velocity}=5.5$ m/s

Hope this answer is helpful.

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