Question

A car moves with a speed of 36km/hr, increases its speed to 108 km/hr in 8 seconds. Find the force if the mass of the car along with the passengers is 1200 kg.

Answer 1

Given:

Initial velocity of car,u= 36 km/h

Final velocity of car,v= 108 km/h

Time taken by car,t= 8 s

Mass of car along with passengers,m= 1200 kg

To Find:

Force applied by car

Solution:

We know that,

• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

t is time taken

• $\pink{\boxed{\bf{Force=Mass\times Acceleration}}}$

$\rule{190}{1}$

It is given that,

$\longrightarrow\rm{u=36\ km/h=36\times\dfrac{5}{18}\ m/s=10\ m/s}$

$\longrightarrow\rm{v=108\ km/h=108\times\dfrac{5}{18}\ m/s=30\ m/s}$

$\rule{190}{1}$

Let the acceleration of car be a

So, on applying first equation of motion on car,we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{30=10+a(8)}$

$\longrightarrow\rm{30-10=8a}$

$\longrightarrow\rm{20=8a}$

$\longrightarrow\rm{8a=20}$

$\longrightarrow\rm{a=\dfrac{20}{8}}$

$\longrightarrow\rm{a=2.5\ m/s^{2}}$

$\rule{190}{1}$

Let the force applied by car be F

So,

$\longrightarrow\rm{F=ma}$

$\longrightarrow\rm{F=1200\times2.5}$

$\longrightarrow\rm\green{F=3000\ N}$

Hence, the force applied by car is 3000N.

Answer 2

◘ Given Data ◘

→ Initial velocity, u = 36 km/h

36 km/h = 36 × (5/18) = 5 × 2 = 10 m/s

→ Final velocity, v = 108 km/h

108 km/h = 108 × (5/18) = 5 × 6 = 30 m/s

→Time, t = 8 secs

→Mass, m = 1200 kg

◘ Objective ◘

To find the force(F).

◘ Calculation ◘

We know,

$\rm{a=\dfrac{v-u}{t} }$

Calculating further :-

$\rm{a=\dfrac{30-10}{8} }\\\\\rm{\to a=\dfrac{20}{8} }\\\\\rm{\to a=\dfrac{5}{2} }\\\\\boxed{\rm{\to a=2.5\ m/s^2 }}$

_________________

We also know that,

$\rm{F=m \times a}$

Calculating further :-

$\rm{\to F=1200 \times 2.5}\\\\\rm{\to F=1200 \times \dfrac{25}{10} }\\\\\rm{\to F=120 \times 25}\\\\\boxed{\rm{\color{magenta}{\to 3000\ N}}}$

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