Question #157475

1. The human body can survive an acceleration of less than 250 m/s2. If a person is involved in a car accident with a velocity of 110 km/h and is stopped by an airbag, over what distance must the air bag stop the person to survive the crash.


2. Suppose your electric bicycle (e-bike) is at rest and can reach a velocity of 12 m/s in 9s Assuming that it accelerates at a constant rate; what is its acceleration in m/s2?


3. Consider a jeepney at rest in the terminal accelerates at a constant acceleration of 4.5m/s^2

a. what is the velocity after 7?

b.) how much distance did the jeepney travel at this time


1
Expert's answer
2021-01-22T09:56:19-0500

1)

v2=v02+2ad,v^2=v_0^2+2ad,d=v2v022a,d=\dfrac{v^2-v_0^2}{2a},d=0(110 kmh1000 m1 km1 h3600 s)22(250 ms2)=1.86 m.d=\dfrac{0-(110\ \dfrac{km}{h}\cdot\dfrac{1000\ m}{1\ km}\cdot\dfrac{1\ h}{3600\ s})^2}{2\cdot(-250\ \dfrac{m}{s^2})}=1.86\ m.

2)

a=vv0t=12 ms9 s=1.33 ms.a=\dfrac{v-v_0}{t}=\dfrac{12\ \dfrac{m}{s}}{9\ s}=1.33\ \dfrac{m}{s}.

3)

a)

v=at=4.5 ms27 s=31.5 ms.v=at=4.5\ \dfrac{m}{s^2}\cdot7\ s=31.5\ \dfrac{m}{s}.

b)

d=v0t+12at2=124.5 ms2(7 s)2=110.25 m.d=v_0t+\dfrac{1}{2}at^2=\dfrac{1}{2}\cdot4.5\ \dfrac{m}{s^2}\cdot(7\ s)^2=110.25\ m.


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