A car travelling at a speed of 29.8 km hr-1 brakes and travels a distance of 0.02km before it comes to a complete stop. Find the time taken for the car to come to rest
u=29.8 km/hv=0 km/hs=0.02 kmu=29.8\space km/h\\v=0\space km/h\\s=0.02\space kmu=29.8 km/hv=0 km/hs=0.02 km
From Newton's laws of motion,
⇒v2−u2=2as⇒a=−u22s⇒a=−22,201 km/h2\Rightarrow v^2-u^2=2as\\\Rightarrow a=\dfrac{-u^2}{2s}\\\Rightarrow a=-22,201\space km/h^2⇒v2−u2=2as⇒a=2s−u2⇒a=−22,201 km/h2
v=u+atv=u+atv=u+at
⇒t=−ua=0.00134 hr=4.83 s\Rightarrow t=\dfrac{-u}{a}=0.00134\space hr=4.83\space s⇒t=a−u=0.00134 hr=4.83 s
t = 4.83 s
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment