Question

Starting from rest a car acquired a velocity of 16m/s in 4s.If mass of the cart is 3kg Find Force acting on it.Also find Distance travelled by the car .​

Answer 1

Answer:

Distance covered by car is 32 metres.

Force acting on it is 12 newton.

Explanation:

According to the Question

It is given that

Initial velocity ,u = 0m/s

Final velocity ,v = 16m/s

Time taken ,t = 4s

Mass of cart ,m = 3kg

we need to calculate the force acting on it and distance covered by it .

Firstly we calculate the acceleration of the car .

Using Kinematics Equation

v = u + at

Putting all the value we get

➺ 16 = 0 + a×4

➺ 16 = 4×a

➺ a = 16/4

➺ a = 4m/s²

So, the acceleration of the car is 4m/s².

As we know that Force is calculated by the product of mass and acceleration .

F = ma

substitute the value we get

➺ F = 3×4

➺ F = 12N

Hence, the force acting is 12 newton.

Now, calculating the distance covered by car in given condition.

again using Kinematics Equation

v² = u² + 2as

substitute the value we get

➺ 16² = 0² + 2×4 ×s

➺ 256 = 0 + 8 × s

➺ 256 = 8×s

➺ s = 256/8

➺ s = 32m

Hence, the distance covered by the car is 32 metres.

Step-by-step explanation:

hopes it's helps you :)

Answer 2

Given : Starting from rest a car acquired a velocity of 16m/s in 4s & If mass of the cart is 3kg.

To Find : Find the distance covered by the car ?

____________________________

Solution : Let the force and distance be x.

$~$

$\underline{\frak{As ~we ~know~ that~:}}$

• $\underset{\blue{\sf 1st~ Equation~of\ Motion}}{\underbrace{\boxed{\frak{\pink{v~=~u~+~at}}}}}$

$~$

$\pmb{\sf{\underline{According~ to ~the ~Given ~Question~:}}}$

$~$

Where,

• u = Initial velocity, 0m/s
• v = Final velocity, 16m/s
• t = Time taken, 4s
• m = Mass of cart, 3kg

$~$

◗We need to calculate the force acting on it and distance covered by,

◗Firstly we calculate the acceleration of the car.

$~$

Using formula we get :

• v = u + at

$~$

$\pmb{\sf{\underline{Substituting ~the ~putting ~values~:}}}$

$~$

$\qquad{\sf:\implies{16~=~0~+~a~×~4}}$

$\qquad{\sf:\implies{16~=~4~×~a}}$

$\qquad{\sf:\implies{a~=~\cancel\dfrac{16}{4}}}$

$\qquad:\implies{\underline{\boxed{\frak{\purple{\pmb{a~=~4~m/s^2}}}}}}$★

$~$

Therefore,

• The acceleration of the car is 4m/s².

$~$

$\underline{\frak{As ~we ~know~ that~:}}$

• $\underset{\blue{\sf Force\ Formula}}{\underbrace{\boxed{\frak{\purple{F~=~ma}}}}}$

$~$

$\pmb{\sf{\underline{Substituting ~the ~Given ~values~:}}}$

$~$

• ${\sf\leadsto{F~=~3~×~4}}$
• ${\sf\leadsto{F~=~12N}}$

$~$

Therefore,

• The force acting is 12 Newton.

$~$

Now, calculating the distance covered by car in given condition,

$~$

• Using Kinetics Equation :

$~$

$\underline{\frak{As ~we ~know~ that~:}}$

• $\underset{\blue{\sf Kinetic\ Equation}}{\underbrace{\boxed{\frak{\purple{v^2~=~u^2~+~2as}}}}}$

$~$

$\pmb{\sf{\underline{Substituting ~the ~Given ~values~:}}}$

$~$

$\qquad{\sf:\implies{16^2~=~0^2~+~2~×~4~×~s}}$

$\qquad{\sf:\implies{256~=~0~+~8~×~s}}$

$\qquad{\sf:\implies{256~=~8~×~s}}$

$\qquad{\sf:\implies{s~=~\cancel\dfrac{256}{8}}}$

$\qquad:\implies{\underline{\boxed{\frak{\pink{\pmb{s~=~32~m}}}}}}$★

$~$

Hence,

$\therefore\underline{\sf{The ~distance ~covered ~by~ the~ car ~is~\bf{\underline{\pmb{32~ metres}}}}}$