Question

Find acceleration and distance travelled in 6 sec, when a net force of 75000N accelerates a car of mass 1500kg uniformly from rest.

Answer 1

$\boxed{\bf{ \blue{\bigstar Find \longrightarrow }}}$

Distance Travelled

Acceleration

$\boxed{\bf{\red{\bigstar Given \longrightarrow }}}$

Force = 75000N

Mass = 1500kg

Time = 6 sec

$\boxed{\bf{\orange{\bigstar Formula \ used \longrightarrow }}}$

$\bf{\green{ F=m \times a}}$

where F= Force

m= mass

a= acceleration

$\bf{\green{ s= ut + \dfrac{1}{2} at^2 }}$

where,

s= distance travelled

u= initial velocity

a= acceleration

t= Time

$\boxed{\bf{\pink{\bigstar SoLution \longrightarrow }}}$

Firstly we will find the acceleration->

According to second law of motion,

F=ma

Put value in the formula we get,

$\sf{\implies 75000= 1500 \times a}$

$\sf{ a= \dfrac{ 75000}{1500} = \dfrac{ 750}{15} }$

$\sf{\star a= 50 m/s^2 \star }$

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Now,

We will find the Distance travelled->

By using second equation of motion,

$\bf{ s= ut + \dfrac{1}{2} at^2 }$

put values in this formula we get,

$\sf{ \implies s= 0 \times 6 + \dfrac{1}{2} \times 50 \times 6 \times 6 }$

$\sf{ \implies s= 25 \times 36}$

$\sf{ \star s= 900 m \star }$

Hence, Distance travelled = 900m

Acceleration = 50m/s²

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Answer 2

Given

❍ Net force acting = 75000 N

❍ Mass = 1500 kg

❍ Car starts from rest

❍ Time = 6 seconds.

To find

❒ Acceleration of car

❒ Distance travelled by car in 6 s.

Solution

At first to find acceleration we will use Newton's 2nd law of motion :

❒ Force = Mass × Acceleration

✎ 75000 = 1500 × Acceleration

✎ Acceleration = 75000/1500

✎ Acceleration = 50 m/s²

Therefore, acceleration of car = 50 m/s²

Now to find distance travelled, let's find final velocity of car at first.. For that we will use 1st equation of motion :

❒ v = u + at

✎ v = 0 + 50 × 6

✎ v = 0 + 300

✎ v = 300 m/s

Now we will use 3rd equation of motion :

❒ v² - u² = 2as

✎ (300)² - 0² = 2 × 50 × s

✎ 90000 - 0 = 100s

✎ 100s = 90000

✎ s = 90000/100

✎ s = 900 m

Therefore, distance travelled by car = 900 m