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A car with weight of 7,656newton is moving with 43.13km/h. The driver applies thebreaks to stop the car in 9.38seconds. The magnitude of theforce (in newton) needed tostop the car is:

Sagot :

Givens.

• Weight = 7,656 N.

,

• Initial speed = 43.13 km/h.

,

• Time = 9.38 seconds.

,

• Final speed = 0 km/h. (The car stops)

First, find the acceleration involved.

[tex]\begin{gathered} v_f=v_0+at \\ a=\frac{v_f-v_0}{t} \\ a=\frac{0-43.13\cdot\frac{km}{h}}{9.38\sec } \end{gathered}[/tex]

But, we need to transform the initial speed to meters per second.

[tex]43.13\cdot\frac{km}{h}\cdot\frac{1000m}{1\operatorname{km}}\cdot\frac{1h}{3600\sec}=11.98\cdot\frac{m}{s}[/tex]

Now we can proceed to find the acceleration.

[tex]\begin{gathered} a=\frac{-11.98\cdot\frac{m}{s}}{9.38s} \\ a=-1.28\cdot\frac{m}{s^2} \end{gathered}[/tex]

Once you have the acceleration. Find the mass of the car using the weight formula.

[tex]\begin{gathered} W=mg \\ 7,656N=m\cdot9.8\cdot\frac{m}{s^2} \\ m=\frac{7,656N}{9.8\cdot\frac{m}{s^2}} \\ m=781.22\operatorname{kg} \end{gathered}[/tex]

Then, use Newton's Second Law to find the needed force to stop.

[tex]\begin{gathered} F=ma \\ F=781.22\operatorname{kg}\cdot(-1.28\cdot\frac{m}{s^2}) \\ F=-999.96N \end{gathered}[/tex]

Therefore, the magnitude of the force needed to stop the car is 999.96 Newtons.

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