Question

A boy start towards east with uniform speed 5m/s.
After t = 2 second he turns right and travels 40 m
withsame speed. Again he turns right and travels
for 8second wi th same speed. Find out the
displacement; average speed, average velocity and
total distance travelled

Answer 1

Answer:

2.78 m/s

Explanation:

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The distance covered in the direction of east=d1 =v×t=5×2=10m

The distance covered in first right turn=d2=40m

and the distance covered in second right turn=d3=v×t=5×8=40m

Total distance covered=d=d1+d2+d3=10+40+40=90m

90mTotal displacement= s = s3−s1

$= \sqrt{ {40}^{2} + {30}^{2} } = 50m \\ \\ average \: speed = \frac{total \: distance \: covered}{total \: time \: taken} \\ = \frac{90}{2 + 8 + 8} = \frac{90}{18}$

= 5 m/s

$average \: velocity = \frac{total \: displacement}{total \: time \: taken \: } \\ \\ = \frac{50}{18}$

= 2.78 m/s

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