Answer:
9.63 seconds
Explanation:
we can use the equation of motion under constant acceleration:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Equation of motion}}\\\\v = u + at \\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{ $ v $ iis the final velocity (which is 0 m/s since the car stops),} \\\phantom{ww}\bullet\;\textsf{$ u $ is the initial velocity (which is 35 m/s), } \\\phantom{ww}\bullet\;\textsf{$ a $ is the acceleration, and } \\\phantom{ww}\bullet\;\textsf{$ t $ is the time taken.} \end{array}}[/tex]
We know that the acceleration [tex] a [/tex] can be calculated using Newton's second law:
[tex]\boxed{\begin{array}{l}\underline{\textsf{ Newton's second law}}\\\\F = ma \\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{ $ F $ is the force acting on the object (4000 N in this case),} \\\phantom{ww}\bullet\;\textsf{$ m $ is the mass of the object (1100 kg), and } \\\phantom{ww}\bullet\;\textsf{$ a $ is the acceleration.} \end{array}}[/tex]
So, rearranging the equation to solve for [tex] a [/tex]:
To find out how many seconds it will take for the car to stop,
[tex] a = \dfrac{F}{m} [/tex]
[tex] a = \dfrac{4000 \, \textsf{N}}{1100 \, \textsf{kg}} [/tex]
[tex] a = 3.636 \, \textsf{m/s}^2 [/tex]
Now, we can use the equation of motion to find the time taken ([tex] t [/tex]):
[tex] 0 = 35 \, \textsf{m/s} + (-3.636 \, \textsf{m/s}^2) \times t [/tex]
[tex] 3.636 \, \textsf{m/s}^2 \times t = 35 \, \textsf{m/s} [/tex]
[tex] t = \dfrac{35 \, \textsf{m/s}}{3.636 \, \textsf{m/s}^2} [/tex]
[tex] t \approx 9.6259625962596 \, \textsf{s} [/tex]
[tex] t \approx 9.63 \, \textsf{s (in 2 d.p.)} [/tex]
So, it will take approximately [tex] \boxed{9.63} [/tex] seconds for the car to stop.
