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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?

Sagot :

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:

[tex]v_f=v_0+at_{}[/tex]

Where

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

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

[tex]0=v_0+at[/tex]

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

[tex]-v_0=at[/tex]

Now we divide by the time "t":

[tex]-\frac{v_0}{t}=a[/tex]

Now we substitute the values:

[tex]-\frac{28\frac{m}{s}}{5.8s}=a[/tex]

Solving the operations we get:

[tex]-4.83\frac{m}{s^2}^{}=a[/tex]

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

[tex]F=ma[/tex]

Where:

[tex]\begin{gathered} F=\text{ force} \\ m=\text{ mass} \\ a=\text{ acceleration} \end{gathered}[/tex]

Substituting the value we get:

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

Solving the operations we get:

[tex]F=-6148.59N[/tex]

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.