Question

a bus travel 700 km at acertain speed. if the bus driver reduces the speed of bus 20 km/hr then the bus will take 4 hours more to travel the same distance. then find the initila speed of the bus

Answer 1

very nice question .If the certain is given, then the answer would be easy.

Answer 2

Answer:

Answer:

Given :-

Time taken = 4 sec

Final velocity = 36 km/hr × 5/18 = 10 m/s

Mass = 700 kg

To Find :-

Force applied

Solution :-

Firstly let's find Acceleration by using first equation

As we know that

\huge \fbox{V = u + at}

V = u + at

Here,

V = Final Velocity = 10 m/s

U = Initial velocity = 0 m/s

A = Acceleration = ?

T = Time = 4 sec

\tt \: 10 = 0 + a \times 410=0+a×4

\tt \: 0 - 10 = 4a0−10=4a

\tt \: - 10 = 4a−10=4a

\tt \: a = \dfrac{ - 10}{4} = 2.5a=

4

−10

=2.5

Hence :-

Acceleration of body is 2.5 m/s²

Now,

Let's find Distance travelled by using third Equation

\huge \bf \: {v}^{2} - {u}^{2} = 2asv

2

−u

2

=2as

Here,

V = Final Velocity

U = Initial velocity

A = Acceleration

S = Distance

\tt \: {10}^{2} - {0}^{2} = 2( - 2.5)(s)10

2

−0

2

=2(−2.5)(s)

\tt \: 0 - 100 = - 5s0−100=−5s

\tt \: - 100 = - 5s−100=−5s

\tt \: s \: = \dfrac{ - 100}{ - 5} = 20 \: ms=

−5

−100

=20m

Hence :-

Distance travelled by body is 20 m

Now,

Let's find Force applied

As we know that

F = ma

Here,

F = Force Applied

M = Mass

A = Acceleration

\tt \:F = 700 \times 2.5F=700×2.5

\tt \: F = 1750F=1750

Hence :-

Force applied is 1750 N.