Question

A boy stars towards east with uniform speed 5m/s .after t=2 sec he turns right and travels 40 m with same speed .again he turns right and travels for 8secwith same speed.find out displacement ,average speed ,average velocity and totAldistance travelled

Answer 1

Heya mate ,
here is your answer,
_____________

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

____________________

# nikzz

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Answer 2

Answer:

Step-by-step explanation:

Distance covered in 25=v.t=10m v=5m/s

85=v.t=40m

His path :

Taking A are origin

A≡(0,.0)

D=(−30,−40)

(t=

v

d

​

=

5

40

​

=8s)

(1)Displacement =d(AD)=

40

2

+30

2

​

=

1600+900

​

=

2500

​

=

50m

​

(ii) Total distance =10+40+40=

90m

​

(iii) Avg. speed =

TotalTime

TotalDistance

​

=

2+8+8

90m

​

=

18

90

​

=

5m/s

​

(iv)Avg. velocity =

Totaltime

TotalDisplacement

​

=

18

50

​

=

2.78m/s

​

in

AD

direction

​

solution