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Motion in a Straight Line

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Question : 6 of 27
Marks: +1, -0
A car moving along a straight highway with speed of 126 km h−1^{-1} is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Solution:  
Here, u=126 km h−1u=126\ \text{km}\ \text{h}^{-1}
=126×100060×60 m s−1=\frac{126\times1000}{60\times60}\ \text{m s}^{-1}
=35 m s−1=35\ \text{m s}^{-1}
v=0, S=200 m, a=?v=0,\ S=200\ \text{m},\ a=? and t=?t=?
We know, v2=u2+2aSv^{2}=u^{2}+2 a S
∴ 0=(35)2+2×a×200\therefore\ 0=(35)^{2}+2\times a\times 200
or a=−(35)22×200a=\frac{-(35)^{2}}{2\times200}
=−4916=−3.06 m s−2=\frac{-49}{16}=-3.06\ \text{m s}^{-2}
As, v=u+atv = u + at
∴ 0=35+(−4916)t\therefore\ 0=35+\left(\frac{-49}{16}\right)t
or t=35×1649t=\frac{35\times16}{49}
=807=11.43 s=\frac{80}{7}=11.43\ \text{s}
–ve sign shows that acceleration is negative, which is called retardationi.e. car is uniformally retarded at a=3.06 m s−2.a = 3.06\ \text{m s}^{-2}.
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