Answer:
1) 9.83 m/s (2 d.p.)
2) 0.04 km (2 d.p.)
Explanation:
Constant Acceleration Equations (SUVAT)
[tex]\boxed{\begin{array}{c}\begin{aligned}v&=u+at\\\\s&=ut+\dfrac{1}{2}at^2\\\\ s&=\left(\dfrac{u+v}{2}\right)t\\\\v^2&=u^2+2as\\\\s&=vt-\dfrac{1}{2}at^2\end{aligned}\end{array}} \quad \boxed{\begin{minipage}{4.6 cm}$s$ = displacement in m\\\\$u$ = initial velocity in ms$^{-1}$\\\\$v$ = final velocity in ms$^{-1}$\\\\$a$ = acceleration in ms$^{-2}$\\\\$t$ = time in s (seconds)\end{minipage}}[/tex]
When using SUVAT, assume the object is modeled as a particle and that acceleration is constant.
Part 1
Given:
- u = 22.1 km/h
- a = 0.84 m/s²
- t = 4.4 s
Convert km/h into m/s by dividing by 3.6:
[tex]\implies \sf 22.1\;km/h=\dfrac{22.1}{3.6}\;m/s[/tex]
Substitute the given values into the formula and solve for v:
[tex]\begin{aligned}v&=u+at\\\implies v & = \dfrac{22.1}{3.6}+0.84(4.4)\\v & = 6.1388...+3.696\\v & = 9.834888...\\v & = 9.83 \sf \;m/s\;(2\:d.p.)\end{aligned}[/tex]
Therefore, the final speed of the car was 9.83 m/s (2 d.p.).
Part 2
Substitute the given values into the formula and solve for s, remembering to use the initial velocity in m/s:
[tex]\begin{aligned}s & = ut+\dfrac{1}{2}at^2\\\implies s &=\dfrac{22.1}{3.6}(4.4)+\dfrac{1}{2}(0.84)(4.4)^2\\s &=27.0111...+8.1312\\s &=35.1423111...\sf m\end{aligned}[/tex]
Convert meters into kilometers by dividing by 1000:
[tex]\implies \dfrac{35.1423111...}{1000}=0.0351423111...=0.04 \sf \; km\;(2\:d.p.)[/tex]
Therefore, the displacement of the car after 4.4 s was 0.04 km (2 d.p.).
