Question

in covering a certain distance the time decreases by 20% if speed (uniform) increases by 20 m/s what is the original speed

Answer 1

Answer:

Original speed = 80 m/s

Explanation:

Let original speed = $x \ m/s$

Let time taken = $y\ s$

Formula for distance is:

$\text{Distance = Speed }\times \text{Time}$

So, distance  $D = x \times y ...... (1)$

It is given that Speed is increased by 20%,

New Speed = $(x + 20) \ m/s$

New Time = $\frac{4}{5} y$

Distance , $D = (x + 20) \times \dfrac{4}{5}y ...... (2)$

Equating (1) and (2):

$xy = (x + 20) \times \dfrac{4}{5}y\\\Rightarrow xy = \dfrac{4}{5}xy + 20\times \dfrac{4}{5}y\\\Rightarrow xy = \dfrac{4}{5}xy + 16y\\\Rightarrow \dfrac{1}{5}xy = 16y\\\Rightarrow x = 80\ m/s$

Hence, Original speed = 80 m/s

Answer 2

Explanation:

keeping the distance constant and changing the values of time and speed

Hope it helps..!