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Keith_Richards [23]
3 years ago
11

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?
Physics
1 answer:
Llana [10]3 years ago
3 0

Answer:

  • a = - 8.617 \frac{m}{s^2}
  • a =  - 0.8798 \ g
  • Tb = 6.64 \ s
  • \frac{Tb}{Tr} = 16.6

Explanation:

<h3>a and b</h3>

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

<h3>c</h3>

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

<h3>d</h3>

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

\frac{Tb}{Tr} = 16.6

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