Question

a person travels 500m with the speed of 54km/hr and then rereturns back taking 25s. Find the average speed​

Answer 1

Given :

A person travels 500m with the speed of 54km/hr then returns back taking 25s.

To Find :

Average speed of the person.

Solution :

Average speed is defined as the ratio of total distance travelled to the total time taken.

• It is a scalar quantity having only magnitude.
• It can't be negative or zero.
• SI unit : m/s
• Dimension formula : [L¹T‾¹]

$\sf:\implies\:V_{av}=\dfrac{d_1+d_2}{t_1+t_2}$

We know that time is measured as the ratio of distance travelled to the speed.

• 54 km/hr = 15 m/s

$\sf:\implies\:V_{av}=\dfrac{500+500}{(500/15)+25}$

$\sf:\implies\:V_{av}=\dfrac{1000}{33.33+25}$

$\sf:\implies\:V_{av}=\dfrac{1000}{58.33}$

$:\implies\:\underline{\boxed{\bf{\orange{V_{av}=17.14\:ms^{-1}}}}}$

Answer 2

Given :

A person travels 500m with the speed of 54km/hr then returns back taking 25s.

To Find :

Average speed of the person.

Solution :

Average speed is defined as the ratio of total distance travelled to the total time taken.

It is a scalar quantity having only magnitude.

It can't be negative or zero.

SI unit : m/s

Dimension formula : [L¹T‾¹]

$\sf:\implies\:V_{av}=\dfrac{d_1+d_2}{t_1+t_2}$

We know that time is measured as the ratio of distance travelled to the speed.

54 km/hr = 15 m/s

$\sf:\implies\:V_{av}=\dfrac{500+500}{(500/15)+25}$

$\sf:\implies\:V_{av}=\dfrac{1000}{33.33+25}$

$\sf:\implies\:V_{av}=\dfrac{1000}{58.33}$

$:\implies\:\underline{\boxed{\bf{\orange{V_{av}=17.14\:ms^{-1}}}}}$