Question

Starting from a stationary position a bus attains a velocity of 6m/s in 30s .Then the driver of the bus applies a brake such that the velocity of the bus comes down to 4m/s in the next 5s .Cal culate the arceleration of the bus in both cases.​

Answer 1

Explanation:

For first case:

Initial velocity =0

Final velocity =60m/s

Time =30sec

v=u+at

6=0+a×30

6=30a

a=

30

6

=0.2

a=0.2m/s

2

For second case:

Initial velocity =6m/s

Final velocity =4m/s

Time taken =5sec

v=u+at

4=6+a×5

4−6=5a

−2=5a

a=−

5

2

a=−0.4m/s

2

Answer 2

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CASE 1 :-

Bus is starting from rest.

Attains a velocity of 6m/sec.

Time taken = 30sec.

• u = 0m/sec

• v = 6m/sec

• t = 30sec

• a = ?

By 1st Kinematical eqⁿ,

$\\ \bigstar \: \boxed{ \bf \: v = u + at}$

➤ Putting values ,

➛ 6 = 0 + a(30)

➛ 6 = 30a

➛ a = 6/30

➛ a = 0.2 m/sec²

$\\ \therefore \rm \: \pink{ acceleration \: is \: 0.2m {s}^{ - 2} }$

_______________________

CASE 2 :-

Final velocity of CASE 1 will be initial velocity here.

Final velocity = 4m/sec

Time taken = 5sec

• u = 6m/sec

• v = 4m/sec

• t = 5sec

• a = ?

We know ,

v = u + at

➤ Putting values ,

➛ 4 = 6 + a(5)

➛ 4 - 6 = 5a

➛ -1 = 5a

➛ a = -1/5

➛ a = -0.2m/sec²

$\\ \therefore \rm{ \pink{ acceleration \: is \: - 0.2 {ms}^{ - 2} }}.$