A car of mass 1100 kg that is travelling at 27 m/s starts to slow down and comes to a complete stop in 578 m. The magnitude of the average braking force acting on the car is 693 N.
Explanation:
The given car of mass of 1100 kg has been travelling at a speed of 27 m/s and the breaking force was applied. To determine the breaking force, following steps can be undertaken,
Applying the equations of motion, as we know,
[tex]v^{2}=u^{2}-2 a s[/tex]
As the car comes to rest, final velocity v=0. Substituting the known value into the third equation of motion as mentioned above, we get,
[tex]0=(27 \times 27)-(2 a \times 578)[/tex]
[tex]2 a \times 578=27 \times 27[/tex]
Then, Deacceleration is
[tex]a=\frac{27 \times 27}{2 \times 578}=\frac{729}{1156}=0.63 \mathrm{m} / \mathrm{s}^{2}[/tex]
Now to calculate the force,
[tex]F=m a=1100 \times 0.63=693 \mathrm{N}[/tex]
A car of mass 1100 kg that is traveling at 27 m/s starts to slow down and comes to a complete stop in 578 m, experiences an average braking force of -6.9 × 10² N.
A car is initially moving at 27 m/s (u). After a displacement of 578 m (s) it stops, that is, its final velocity (v) is zero. We can calculate the acceleration (a) of the braking force using the following kinematic equation.
[tex]v^{2} = u^{2} + 2 \times a \times s\\\\a = \frac{v^{2} - u^{2}}{2 \times s} = \frac{(0m/s)^{2} - (27m/s)^{2}}{2 \times 578m} = -0.63 m/s^{2}[/tex]
The car of mass (m) 1100 kg experiences an acceleration of -0.63 m/s². We can calculate the associated force (F) using Newton's second law of motion.
[tex]F = m \times a = 1100 kg \times (-0.63 m/s^{2} ) = -6.9 \times 10^{2} N[/tex]
A car of mass 1100 kg that is traveling at 27 m/s starts to slow down and comes to a complete stop in 578 m, experiences an average braking force of -6.9 × 10² N.
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