Respuesta :
Answer:8.75 s,
136.89 m
Explanation:
Given
Initial velocity[tex]=70 mph\approx 31.29 m/s[/tex]
velocity after 5 s is [tex]30 mph\approx 13.41 m/s[/tex]
Therefore acceleration during these 5 s
[tex]a=\frac{v-u}{t}[/tex]
[tex]a=\frac{13.41-31.29}{5}=-3.576 m/s^2[/tex]
therefore time required to stop
v=u+at
here v=final velocity =0 m/s
initial velocity =31.29 m/s
[tex]0=31.29-3.576\times t[/tex]
[tex]t=\frac{31.29}{3.576}=8.75 s[/tex]
(b)total distance traveled before stoppage
[tex]v^2-u^2=2as[/tex]
[tex]0^2-31.29^2=2\times (-3.576)\cdot s[/tex]
s=136.89 m
A. The time taken for the car to stop is 8.75 s
B. The distance travelled when the brakes were applied till the car stops is 136.89 m
A. Determination of the time taken for the car to stop.
- We'll begin by calculating the deceleration of the car
Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s
Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s
Time (t) = 5 s
Deceleration (a) =?
[tex]a \: = \frac{v \: - u}{t} \\ \\ a = \frac{13.41 - 31.29}{5} \\ \\ a \: = \frac{ - 17.88}{5} \\ \\ [/tex]
a = –3.576 m/s²
- Finally, we shall determine the time taken for the car to stop.
Initial velocity (u) = 31.29 m/s
Final velocity (v) = 0 m/s
Deceleration (a) = –3.576 m/s²
Time (t) =?
[tex]v \: = u \: + at \\ 0 \: = 31.29 \: + \: ( - 3.576 \times t) \\ 0 \: = 31.29 \: - 3.576 \times t \\ collet \: like \: terms \\ 0 - 31.29 \: = - 3.576 \times t \\ - 31.29 \: = - 3.576 \times t \\ divide \: both \: side \: by \: - 3.576 \\ t \: = \frac{- 31.29}{- 3.576} \\ [/tex]
t = 8.75 s
Thus, the time taken for the car to stop is 8.75 s
B. Determination of the total distance travelled when the brakes were applied.
Initial velocity (u) = 31.29 m/s
Final velocity (v) = 0 m/s
Deceleration (a) = –3.576 m/s²
Distance (s) =?
[tex]{v}^{2} = {u}^{2} + 2as \\ {0}^{2} = {31.29}^{2} + (2 \times - 3.576 \times s) \\ 0 = 979.0641 - 7.152 s \\ collect \: like \: terms \\ 0 - 979.0641 = - 7.152 s \\ - 979.0641 = - 7.152 s \\ divide \: both \: side \: by \: - 7.152 \\ s = \frac{- 979.0641}{- 7.152} \\ \\ [/tex]
s = 136.89 m
Therefore, the total distance travelled by the car when the brakes were applied is 136.89 m
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