Question

A car of mass 1000kg and a bus of mass 8000kg are moving with same velocity of 36km h*-1.find the forces tostop both the car and the bus in 5s.

Plzzzz tell.....will mark as BRAINLIEST.......

Answer 1

Given:-

Mass of car = 1000kg

Mass of bus = 8000kg

Time taken by both car and bus = 5 seconds

Initial velocity of both = 36km/h

Final velocity (v) = 0

To find :-

The force exerted on both car and bus

• First find the acceleration ,By 1st equation of motion

$\boxed{\sf{ v= u+at}}$

$\sf u= \cancel{36}\times \dfrac{5}{\cancel{18}}= 5\times 2\\ \\ \longmapsto\sf u= 10m/s$

$\implies\sf 0 = 10+a\times 5\\ \\ \implies\sf -10= 5a\\ \\ \implies \sf \cancel{\dfrac{-10}{5}}=a\\ \\ \implies\sf a= -2m/s^2$

• The acceleration of both car and bus will be same
• Now find the force applied on it

$\boxed{\sf{Force (f) = Mass(m) \times Acceleration (a)}}$

• Force applied on car

$\dashrightarrow\sf f= m\times a\\ \\ \dashrightarrow\sf f= 1000\times -2 \\ \\ \dashrightarrow\boxed{\sf{f= -2000N}}$

• Now force applied on bus acceleration will remain same

$\dashrightarrow\sf f= m\times a\\ \\ \dashrightarrow\sf f= 8000\times -2 \\ \\ \dashrightarrow\boxed{\sf{f= -16000N}}$

◆ Force applied on car = (-2000N)

◆ Force applied on bus = (-16000N)

Answer 2

Answer:

Mass of car = 1000kg

Mass of bus = 8000kg

Time taken by both car and bus = 5 seconds

Initial velocity of both = 36km/h

Final velocity (v) = 0

To find :-

The force exerted on both car and bus

First find the acceleration ,By 1st equation of motion

\boxed{\sf{ v= u+at}}

v=u+at

\begin{gathered}\sf u= \cancel{36}\times \dfrac{5}{\cancel{18}}= 5\times 2\\ \\ \longmapsto\sf u= 10m/s \end{gathered}

u=

36

×

18

5

=5×2

⟼u=10m/s

\begin{gathered}\implies\sf 0 = 10+a\times 5\\ \\ \implies\sf -10= 5a\\ \\ \implies \sf \cancel{\dfrac{-10}{5}}=a\\ \\ \implies\sf a= -2m/s^2\end{gathered}

⟹0=10+a×5

⟹−10=5a

⟹

5

−10

=a

⟹a=−2m/s

2

The acceleration of both car and bus will be same

Now find the force applied on it

\boxed{\sf{Force (f) = Mass(m) \times Acceleration (a)}}

Force(f)=Mass(m)×Acceleration(a)

Force applied on car

\begin{gathered}\dashrightarrow\sf f= m\times a\\ \\ \dashrightarrow\sf f= 1000\times -2 \\ \\ \dashrightarrow\boxed{\sf{f= -2000N}}\end{gathered}

⇢f=m×a

⇢f=1000×−2

⇢

f=−2000N

Now force applied on bus acceleration will remain same

\begin{gathered}\dashrightarrow\sf f= m\times a\\ \\ \dashrightarrow\sf f= 8000\times -2 \\ \\ \dashrightarrow\boxed{\sf{f= -16000N}}\end{gathered}

⇢f=m×a

⇢f=8000×−2

⇢

f=−16000N

◆ Force applied on car = (-2000N)

◆ Force applied on bus = (-16000N)