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How much force is required to bring a car of mass 1,273 kg, initially traveling at 28 m/s, to rest in 5.8 s?

User Wardk
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1 Answer

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We are asked to determine the force required to stop a car traveling at a speed of 28 m/s in 5.8s. To do that we will calculate the acceleration of the car first. The acceleration is determined using the following equation:


v_f=v_0+at_{}

Where


\begin{gathered} v_f,v_0=\text{ final and initial velocities} \\ a=\text{ acceleration} \\ t=\text{ time} \end{gathered}

Since we are calculating the force when the car stops this means that the final velocity is zero:


0=v_0+at

Now we solve for the acceleration "a", first by subtracting the initial velocity from both sides:


-v_0=at

Now we divide by the time "t":


-(v_0)/(t)=a

Now we substitute the values:


-(28(m)/(s))/(5.8s)=a

Solving the operations we get:


-4.83(m)/(s^2)^{}=a

Now, we use Newton's second law to determine the force:


F=ma

Where:


\begin{gathered} F=\text{ force} \\ m=\text{ mass} \\ a=\text{ acceleration} \end{gathered}

Substituting the value we get:


F=(1273\operatorname{kg})(-4.83(m)/(s^2))

Solving the operations we get:


F=-6148.59N

Therefore, the required force is -6148.59 Newtons. The negative sign means that the force is acting in the opposite direction of the movement of the car.

User Soupette
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