Answer:
[tex]2.5\; {\rm m\cdot s^{-1}}[/tex].
Approximately [tex]16\; {\rm m}[/tex].
Explanation:
When acceleration is constant, the SUVAT equations would apply. Let [tex]a[/tex] denote acceleration, [tex]u[/tex] denote initial velocity, [tex]v[/tex] denote velocity after acceleration, [tex]x[/tex] denote displacement, and [tex]t[/tex] denote the duration of acceleration.
- [tex]v = u + a\, t[/tex] relates the velocity after acceleration to the duration of the acceleration.
- [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] relates the displacement after acceleration to initial velocity, final velocity, and acceleration.
Right before braking, the vehicle (initial velocity: [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) would have accelerated at [tex]a = 1.3\; {\rm m\cdot s^{-2}}[/tex] for [tex]t = 4.0\; {\rm s}[/tex]. Apply the equation [tex]v = u + a\, t[/tex] to find the velocity of the vehicle after this period of acceleration:
[tex]\begin{aligned} v &= u + a\, t \\ &= (0\; {\rm m\cdot s^{-1}}) + (1.3\; {\rm m\cdot s^{-2}})\, (4.0\; {\rm s}) \\ &= 5.2\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Apply the equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the displacement [tex]x[/tex] of this vehicle at that moment:
[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{(5.2\; {\rm m\cdot s^{-1}})^{2} - (0\; {\rm m\cdot s^{-1}})^{2}}{2 \times 1.3\; {\rm m\cdot s^{-2}}} \\ &= 10.4\; {\rm m}\end{aligned}[/tex].
It is given that acceleration is constant (at [tex]a = (-1.8\; {\rm m\cdot s^{-2}})[/tex]) during braking. The velocity before this period of acceleration (initial velocity) would now be [tex]u = 5.2\; {\rm m\cdot s^{-1}}[/tex]. After [tex]t = 1.50\; {\rm s}[/tex] of braking, the velocity of this vehicle would be:
[tex]\begin{aligned} v &= u + a\, t \\ &= (5.2\; {\rm m\cdot s^{-1}}) + (-1.8\; {\rm m\cdot s^{-2}})\, (1.50\; {\rm s}) \\ &= 2.5\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Apply the equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the displacement of this vehicle during this [tex]t = 1.50\; {\rm s}[/tex] of braking. Note that this displacement gives the position relative to where this vehicle started braking, not where it started from rest.
[tex]\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{(2.5\; {\rm m\cdot s^{-1}})^{2} - (5.2\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-1.8)\; {\rm m\cdot s^{-2}}} \\ &= 5.775\; {\rm m}\end{aligned}[/tex].
The total displacement of this vehicle (relative to where it started from rest) is the sum of the displacement during acceleration and the displacement during braking: [tex]10.4\; {\rm m} + 5.775\; {\rm m} \approx 16\; {\rm m}[/tex]. In other words, after that [tex]t = 1.50\; {\rm s}[/tex] of braking this vehicle would have been approximately [tex]16\; {\rm m}[/tex] from where it started.
