Question

a rally car accelerates from 100km/h to 150km/h in 5s
what is its acceleration in
km/h/s
m/s^2

Answer 1

Answer:

In m/s²

$acceleration(a) = 2.78m/s {}^{2}$

Im km/hr

$acceleration(a) = 35714.28km/h {}^{2}$

Explanation:

Given values:-

• Final velocity (Vf)= 150km/h
• Initial velocity (Vi)=100km/h
• Time(t)=5sec

Thus,

• 1sec = 0.000278hr
• 5sec = 0.00138889hr

where as also

• Vf = 41.7m/s
• Vi=27.7m/s

required value:-

• acceleration (a) =?

solution/work-out:-

Firstly,recall the Velocity-Time equation:

$vf = vi + at$

substitute known variables into the equation-

$(41.6) = (27.7) + a(5)$

solve for acceleration

$a = 2.78m/s {}^{2}$

Fairly use the same method to solve in km/hr

Hope it helps !!!

Answer 2

Answer:

10 km/h/s

2.78 m/s²  (2 d.p.)

Explanation:

$\boxed{\begin{minipage}{9 cm}\underline{The Constant Acceleration Equations (SUVAT)}\\\\s = displacement in m (meters)\\u = initial velocity in m s ^{-1} (meters per second)\\v = final velocity in m s ^{-1} (meters per second)\\a = acceleration in m s ^{-2} (meters per second per second)\\t = time in s (seconds)\\\\When using SUVAT, assume the object is modeled\\ as a particle and that acceleration is constant.\end{minipage}}$

Acceleration in km/h/s

Given:

• u = 100 km/h
• v = 150 km/h
• t = 5 s

$\begin{aligned}\textsf{Using }v & = u+at\\\implies 150 & = 100+5a\\50 & = 5a\\a & = 10\:\sf km/h/s\end{aligned}$

Acceleration in m/s²

As:

• 1000 m = 1 km
• 3600 s = 1 hour

Therefore:

$\implies \sf 100\: km/h = \dfrac{100 \times 1000}{3600}= \dfrac{250}{9}\: m/s$

$\implies \sf 150\: km/h = \dfrac{150\times 1000}{3600}= \dfrac{125}{3}\: m/s$

Given:

• u = 250/9 m/s
• v = 125/3 m/s
• t = 5 s

$\begin{aligned}\textsf{Using }v & = u+at\\\implies \dfrac{125}{3}& = \dfrac{250}{9}+5a\\\dfrac{125}{9} & = 5a\\a & = \dfrac{25}{9}\\ & = 2.78\:\sf m/s^2\:\:(2\:d.p.)\end{aligned}$