A truck driver is driving at a velocity of 60 km/h. He sees a danger ahead and steps on his brakes. He stop the truck in exactly 5 seconds. How far does he go before coming to a stop?
v0=60kmh=60∗10003600≈16.7msv_0 = 60\frac{km}{h}=\frac{60*1000}{3600}\approx16.7\frac{m}{s}v0=60hkm=360060∗1000≈16.7sm
v1=0v_1 = 0v1=0
t=5st = 5st=5s
a=v1−v0t=−3.34ms2a = \frac{v_1-v_0}{t}= -3.34\frac{m}{s^2}a=tv1−v0=−3.34s2m
s=v0t+at22=16.7∗5−3.34∗252=41.75ms = v_0t+\frac{at^2}{2}=16.7*5- \frac{3.34*25}{2}= 41.75ms=v0t+2at2=16.7∗5−23.34∗25=41.75m
Answer: 41.75m\text{Answer: } 41.75mAnswer: 41.75m
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