Question

Heya!

Solve these numericals :

1》A car is traveling at 54 km/hr. If it's velocity increases to 72 km/hr in 5 second, then find the acceleration of the car.

2》 A train takes 2 hours to reach station B from station A and then 3 hours to return from B to A. The distance between the two stations is 200 km. Find the average speed & velocity of the train.

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Answer 1

hey
here is ur answer
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1)➡u=54km/h=54×5/18=15m/s
v=72km/h=72×5/18=20m/s
t=5 sec
we know that
a=v-u/t
a=(20-15)/5
a=(5)/5
a=1m/s^2

2)➡A train takes 2 hours to reach station b from 
station a,
and then 3 hours to return from station b to station a.
The distance between the stations is 200 km.
Find the (a) average speed 
let a = average speed, it took 5 hrs to travel 400 km
a=400/5
a=80km/hr
(b) average velocity of the train.
Velocity involves direction and displacement
Since it ended up in the same place it started (A); velocity is 0  h

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I hope this will help
#Prem✴✌✴✌✴✌✴

Answer 2

Heya user☺☺

Solution 1 - Let the acceleration be a
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So, by the definition, we have

a = (v-u)/t , where v is final velocity and u is initial velocity and t is time

Now, 54km/hr = 54 × 5/18 = 10m/s .......(1)

Also, 72km/hr = 72 × 5/18 = 15m/s .........(2)

so, a = {(2) - (1)}/ t

a = (15 - 10)/5

a = 5/5 m/s^2

a = 1 m/s^2

So, acceleration will be 1 m/s^2 .

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Solution 2 - Let the average speed of the train be x km/hr

Now we know that ,

Average speed = Total distance / Total time

= ( 2 × 200 )/ (2 + 3)

= 400 / 5

= 80 km / hrs.

Now , we know that ,

Velocity of the train = Total displacement / Total time

Here, displacement will be zero as the final position and initial position are same.

So,

Velocity = 0 / 5

= 0 km/hr.

So,

Average speed will be 80 km/hr

And, velocity will be 0 km/hr.

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Hope this will help you☺☺