Question

4. A bus moving on a straight road with a speed of 126 km/h brought to rest after 200 m. Calculate
a) acceleration of the bus and
b) time taken by the bus to come to rest.

Answer 1

$<b><u><i> Hey mate!! Here's your answer</i></u></b>$
______________________________

$\huge{Given}$

A bus is moving on a straight road with a speed of 126 km/h. It is brought to rest after travelling 200 m.

$\huge{To\: Find}$

Acceleration (a) of the bus.
Time taken (t) by bus to come to rest.

$\huge{Solution}$

Initial velocity (u)
= 126 km/h
= 126 km/h
= 126×1000 m/60×60 s
= 126000/3600
= 35 m/s

Final velocity (v) = 0 m/s (because it comes to rest.

Distance travelled (s) = 200 m

Now, using the third equation of motion, which relates position and velocity, we have

2as = v² - u²

=> 2 × a × 200 = 0² - 35²

=> 400a = (-1225)

=> a = (-1225/400)

=> a = -3.06

Hence, acceleration is 3.06 m/s². Negative sign demarcates that the bus is coming to rest (deacclerating or retarding).

Now, using the first equation of motion, which relates velocity and time, we have

v = u + at

=> 0 = 35 + (-3.06 × t)

=> 0 = 35 + (-3.06t)

=> 3.06t = 35

=> t = 35/3.06

=> t = 11.4 sec.

Hence, time taken by bus to come at rest = 11.4 secs.

Finally,

$\boxed{\mathcal{\red{Acceleration \: = \: -3.06 \: m/s^2}}}$
$\boxed{\mathcal{\pink{Time\: Taken \: = \: 11.4\: sec}}}$
______________________________

$\huge{\bold{\mathcal{\green{Thank \: you}}}}$

Answer 2

ANS-

$Acceleration= -3.06 seconds$

$Time taken = 11.04 \: seconds</p><p>$

EXPLANATION:-

$Speed \: of \: the \: bus= 126km/hr$

$Distance \: travelled \: to \: come \: to \: rest= 200m$

First we will convert the above data into a single unit.

To convert km /hr to m/sec we multiply by 5/18

$1km/hr=5/18 \: m/sec$

$126km/hr= 5/18×126 \: m/sec$

$= 35 m/sec$

• The initial velocity of the bus was 35m/sec.
• As the bus came to rest the final velocity of the bus is zero.
• 200m is the distance taken to go from 35m/sec to 0 m/sec

Third equation of motion gives us the relation:

$v^2-u^2=2as$

where

• v= Final velocity
• u = Initial velocity
• a= acceleration
• d= distance covered from initial to final velocity.

Applying the above values in the question:

$0^2-35^2= 2×a×200$

$-1225= 400a$

$a= -3.06 \: m/s^2$

Negative sign shows that the bus is coming to rest ie deaccelerating.

Now we have to find the time taken by the body to come to rest.

The first equation of motion gives us the relation:

$v=u+at$

• v= Final velocity
• u= initial velocity
• a= acceleration
• t= time taken coming from inital to final velocity.

Putting the above values here we get

$0=35+(-3.06t)$

$35= 3.06t$

$t= 35/3.06= 11.04 \: seconds$

So time taken by the bus to come to rest is 11.04 seconds and it's acceleration is -3.06 seconds

Further links-

https://www.google.com/url?sa=t&source=web&rct=j&url=https://brainly.in/question/1418900&ved=2ahUKEwjs1v2DuNz5AhVj7zgGHQKHB1MQjjh6BAgcEAE&usg=AOvVaw3wkCzm6c6ktZBOWZepeddh

https://brainly.in/question/17883870

CODE #SPJ5