Question

A car can be braked to a stop from the autobahn-like speed of 206 km/h in 190 m. Assuming the acceleration is constant, find its magnitude in (a) SI units and (b) in terms of g. (c) How much time Tb is required for the braking? Your reaction time Tr is the time you require to perceive an emergency, move your foot to the brake, and begin the braking. If Tr = 400 ms, then (d) what is the ratio Tb/Tr?

Answer 1

Answer:

• $a = - 8.617 \frac{m}{s^2}$

• $a =  - 0.8798 \ g$

• $Tb = 6.64 \ s$

• $\frac{Tb}{Tr} = 16.6$

Explanation:

a and b

We can use the following formula, from kinematics:

$(v_f)^2-(v_i)^2 = 2  \ a \  d$

where $v_f$ is the final speed, $v_i$ is the initial speed, a is the acceleration and d the distance traveled.

Knowing that the final speed is zero, and the initial speed is

$v_i = 206 \frac{km}{h} = 206 \frac{km}{h} \frac{1000 \ m}{1 \ km} \frac{1 \ h}{3600 \ s} = 57.22 \frac{m}{s}$

we obtain

$(0)^2-(57.22 \frac{m}{s})^2 = 2  \ a \  190 \ m$

$a = \frac{ - \ 3,274.13 \frac{m^2}{s^2} }{ 2 \ 190 \ m}$

$a = - 8.617 \frac{m}{s^2}$

Now, knowing that

$g = 9.806 \frac{m}{s^2}$

then

$a = \frac{- 8.617 \frac{m}{s^2}} { 9.806 \frac{m}{s^2} } \ g$

$a = -  0.8798 \ g$

c

We can use the following formula

$v(t) = v_0 + a t$

so

$0 =  57.22 \frac{m}{s} - 8.617 \frac{m}{s^2} \ Tb$

$Tb = \frac{ 57.22 \frac{m}{s} }{ 8.617 \frac{m}{s^2} }$

$Tb = 6.64 \ s$

d

$\frac{Tb}{Tr} = \frac{ 6.64 \ s }{ 0.4 s }$

$\frac{Tb}{Tr} = 16.6$