Question

A man walks at a speed of 1m/s and covers a certain distance in some time. But , if he walks at a speed of 3m/s , he covers an additional distance of 15m in the same time. What is the distance he actually covered?

Answer 1

EXPLANATION.

• GIVEN

man walk at a speed of 1 m/s and covers a certain

distance in some time.

if he walks at a speed of 3 m/s, he covers an

additional distance of 15 m in the same time.

To find what is the distance he

actually covered?

according to the question,

Let

D = distance, S = speed, T = time

in this case time T is the same

Distance = Speed X Time

D = S X T

in case = 1

D = 1 X T = D = T.....(1)

in case = 2

D + 15 = 3 X T .....(2)

From equation (1) and (2) we get,

T + 15 = 3T

2T = 15

T = 7.5 seconds

As, Distance covered in first case =

D = S X T = 1 m/s X 7.5 seconds = 7.5 meters.

Distance covered in second case =

7.5 + 15 = 22.5 meters.

Answer 2

$\bf \huge \red{question : } -$

A man walks at a speed of 1m/s and covers a certain distance in some time. But , if he walks at a speed of 3m/s , he covers an additional distance of 15m in the same time. What is the distance he actually covered?

$\bf \huge \red{answer} : -$

22.5 meters

$\bf \huge \red{solution : - }$

Let Distance=D;

Speed= S and

Time=T

In this case, Time T is the same.

$\bf \huge\fbox \green{Distance = Speed \: \times Time (or )D= S \times T}$

In the first case, D = 1 xT or D =T ——->(1)

In the second case, D + 15 = 3 xT or 3 T ——->(2)

Using value of D as per (1) in (2),

T +15 = 3 T or 2T =15 or T=15/2= 7.5 seconds

As D=S x T, Distance covered in first case = 1m/second x 7.5 second= 7.5 meters

Distance covered in 2nd case = 7.5 + 15 = 22.5