Question

A bus is moving at 10 m/s. Driver increased its speed  to 25 m/s in 5 sec. find average velocity and from that find distance travelled by bus in that 5 seconds.​

Answer 1

Answer:

Average velocity = 17.5 m/s

Distance travelled = 87.5 m

Explanation:

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

• Initial velocity (u) = 10 m/s
• Final velocity (v) = 25 m/s
• Time taken (t) = 5 seconds

We are asked to calculate average velocity and from that distance travelled by bus in that 5 seconds.

Finding average velocity :

We know that,

$\longmapsto \bf { \overline{\nu} = \dfrac{v + u}{2} }$

• $\rm \overline{\nu}$ denotes average velocity
• v denotes final velocity
• u denotes initial velocity

Substituting the values, we have :

$\longmapsto \rm { \overline{\nu} = \dfrac{25 \; ms^{-1} + 10 \; ms^{-1} }{2} }$

$\longmapsto \rm { \overline{\nu} = \dfrac{ 35 \; ms^{-1} }{2} }$

$\longmapsto \bf { \overline{\nu} =17.5 \; ms^{-1} }$

∴ Average velocity of the bus is 17.5 m/s.

Finding distance travelled :

Now, we have to find the distance travelled from average velocity.

As we know that,

⇒ Distance = Average velocity × Time taken

We've already calculated the average velocity that is 17.5 m/s.

⇒ Distance = 17.5 m/s × 5 s

⇒ Distance = 87.5 m

∴ Distance travelled by bus in 5 seconds is 87.5 m.