Question

4. A boy walks a distance of 30 m in 1 minute and
another 30 m in 1.5 minute. Describe the type of
motion of the boy and find his average speed
in ms-1​

Answer 1

Explanation:

As per the provided information in the given question, we have :

• A boy walks a distance of 30 m in 1 minute.
• And, another 30 m in 1.5 minutes.

We are asked to find what type of motion the body has and his average speed in m/s.

→ The motion of the boy is non-uniform motion. It is because the boy is covering equal distances in unequal intervals of time.

Whenever the body covers equal distances in unequal intervals of time and vice-versa. Then its motion is non-uniform.

$\rule{200}2$

Now, we have to find his average speed(in m/s). Average speed refers to the total distance divided by total time.

$\longmapsto \bf { Speed_{(avg)} = \dfrac{Total \; distance}{Total \; time} }$

Since, we have to find average speed in m/s. So, it must be sure that the total distance should be in metre and total time should be in seconds.

Finding total distance :

⇒ Total distance = (30 + 30) m

⇒ Total distance = 60 m

Finding total time :

⇒ Total time = (1 + 1.5) minutes

⇒ Total time = 2.5 minutes

Converting minutes into seconds :-

• 1 minute = 60 seconds

⇒ Total time = (2.5 × 60) seconds

⇒ Total time = 150 seconds

Now, substitute the value of total distance and total time in the formula of average speed.

$\longmapsto \bf { Speed_{(avg)} = \dfrac{Total \; distance}{Total \; time} }$

$\longmapsto \rm { Speed_{(avg)} =\Bigg ( \dfrac{60}{150}\Bigg ) \; ms^{-1} }$

$\longmapsto \rm { Speed_{(avg)} =\Bigg ( \dfrac{6}{15}\Bigg ) \; ms^{-1} }$

$\longmapsto \rm { Speed_{(avg)} =\Bigg ( \dfrac{2}{5}\Bigg ) \; ms^{-1} }$

$\longmapsto \bf { Speed_{(avg)} = 0.4 \; ms^{-1} }$

∴ Average speed is 0.4 m/s.