Question

14.
Kurt is jogging along the outer edge of a park. The dimensions of the park are shown in the
given figure. He takes one minute to complete one revolution around the park.
100 mm
50 m
Som
100 m
Kurt's speed is
A) 1 m/s
B) 5 m/s
C) 50 m/s
D) 100 m/s​

Answer 1

Answer:-

Given:-

• Dimensions of the Park = 100m × 50m
• Time (t) he takes for a complete revolution = 1 mins.

To Find:-

Kurt's speed. (Assume uniform)

__________...

Distance = Perimeter of the park (rect.)

= 2 ( l + b )

= 2 ( 100 + 50 ) m

= 2 • 150 m

= 300 m

Time = 1 mins

We know,

Speed = Distance/Time

i.e. v = s/t

∴ v = 300 m/1mins

➽ v = 300 m/60 s [Since 1 min. = 60s]

➽ v = 5 m/s

∴ Kurt's speed was 5 m s^-1. (B)

Answer 2

Answer:

$Solution -$

$\huge\boxed{speed \: = \: \frac{distance}{time}}$

• As Kurt is jogging along the outer edge of a rectangular park, so his distance will be equal to the perimeter of the rectangular park.

Distance =

$2 \times (l + b) = 2 (100 + 50)$

$2 \times 150 = 300m$

Now,we have

Distance = 300 m

Time = 1 minute = 60 seconds

Speed=?

$\tt\bold\ {speed = \frac{distance}{time}}$

$\tt\ {speed = \frac{300}{60}}$

$\tt\ {speed = \frac{30}{6} = 5m/second}$