Question

A driver accelerated his car first at the rate of 4m/s2 and then at the rate of 8m/s2. Calculate the ratio of the forces exerted by the engines.​

Answer 1

Answer

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Force is that which changes or tries to change the state of rest or of uniform motion of a body in a straight line.

$\boxed{Force = Mass \times Acceleration}$

$\longrightarrow$Mass of the car is same as only one car is accelerated.

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Force when the car accelerates at 4 m/s²

= 4 × m = 4m Newton

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Force when car accelerates at 8 m/s²

= 8 × m = 8m Newton

$\longrightarrow$where m is mass of car.

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Ratio -

$= {\frac{4m}{8m}}$

= 1 : 2

So the ratio of force exerted by engines =

1 : 2

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ADDITIONAL INFORMATION -

$\longrightarrow$Force is an external efferot in form of push or pull which produces or tries to produce acceleration in the body on which it acts.

$\longrightarrow$Inertia - It is the tendency of an object to oppose its rest or state of uniform motion.

$\longrightarrow$Momentum is product of mass and velocity.

$\implies$${\vec{p} = m\vec{v}}$

$\longrightarrow$Change in momentum = Force × time

$\longrightarrow$$Impule = F\Delta t$

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Thanks

Answer 2

✧ Question:

➝ A driver accelerated his car first at the rate of $4ms^{-2}$ and then at the rate of $8ms^{-2}$. Calculate the ratio of the forces exerted by the engines.

✧ To Find :

➝ The Ratio of the forces exerted by the engine.

✧ Given :

• Acceleration $\rightarrow a_{1} = 4ms^{-2}$

• Acceleration $\rightarrow a_{2} = 8ms^{-2}$

✧ We know :

➝ Force = mass × acceleration

$\boxed{\mathtt{F = ma}}$

➝ where,

• F = force exerted
• m = mass
• a = acceleration

✧ Concept :

➝ By individually finding the force in both the cases , and comparing them , we can find the ratio of force exerted by the two engines.

✧ Solution :

Let the mass be m , in both the cases.

☞ Case I ...

➝ Given ,

• $a_{1} = 4ms^{-2}$

• $mass = m$

☞ We Know ,

$\sf{F = ma}$

Putting the value in the formula , we get,

$\Rightarrow F_{1} = m \times 4$

$\Rightarrow F_{1} = 4m\:N$

Force exerted by the engine in $4ms^{-2}$ is $4m\:N$

☞ Case II ...

➝ Given ,

$a_{1} = 8ms{^-2}$

$mass = m$

☞ We Know ,

$\sf{F = ma}$

Putting the value in the formula , we get,

$\Rightarrow F_{1} = m \times 8$

$\Rightarrow F_{1} = 8m\:N$

Force exerted by the engine in $8ms^{-2}$ is $8m\:N$

Ratio of the Forces :

Ratio =

$F_{1} : F_{2}$

$\Rightarrow 4m : 8m$

$\Rightarrow \cancel{4m} : \cancel{8m}$

$\Rightarrow 1 : 2$

Hence , the ratio is 1 : 2.