Question

5. If an object ‘A’ covers a distance of 500m in 10s while an object ‘B’ covers the same
distance in 15s. Calculate the speed of A and B to find out which one is moving faster.

Answer 1

A:-

• Distance=500m
• Time=10s

We know

$\boxed{\sf Speed=\dfrac{Distance}{Time}}$

$\\ \sf\longmapsto Speed=\dfrac{500}{10}$

$\\ \sf\longmapsto Speed=50m/s$

B:-

• Time=15s

$\\ \sf\longmapsto Speed=\dfrac{500}{15}$

$\\ \sf\longmapsto Speed=33.3m/s$

A is moving faster

Answer 2

Answer:

Given,

An object ‘A’ covers a distance of 500m in 10s

So,

$\sf \: S_A = 500 \: m \\ \sf \: t_A = 10 \: s \\ \green{\sf \: velocity \: of \: object (A)\: \: \: V_A = \: to \: find}$

And

A object ‘B’ covers a distance of 500m in 15s

So,

$\sf \:S_B = 500 \: m \\ \sf \: t_B \: = 15 \: s \\ \green{ \sf \: velocity \: of \: object(B) \: \: \: V_B = \: to \: find}$

Finally, we have to find :

☞ Which object is moving faster.

Formula that will be used :

if

distance = S

time = t

Velocity / speed = V

then,

we know that,

$\huge \: \: \: \sf \: S= Vt \\ \implies \huge\: V = \frac{S}{t}$

Solution :

For object 'A' :

$\sf \: V_A = \frac{S_A}{t_A} \\ \: \: \: \: \: \: \: \sf = \frac{500}{10} \\ \: \: \: \: \: \: \sf \: = 50 \: m {s}^{ - 1}$

For object 'B' :

$\sf \: V_B = \frac{S_B}{t_B} \\ \: \sf \: \: \: \: \: \: = \frac{500}{15} \\ \: \: \: \: \: \: \: = \sf 33.33 \: m {s}^{ - 1}$

So we can see that,

$\large \green {\boxed{V_A \: > \: V_B}}$

So, object 'A' is moving faster than object 'B'.