Question

ii). A car moving along a straight road with a
speed of 120 km/hr, is brought to rest by
applying brakes. The car covers a distance
of 100 m before it stops. Calculate
(i) the
average retardation of the car (ii) time
taken by the car to come to rest.
[Ans: 50/9 m/sec, 6 sec)​

Answer 1

$\huge\mathcal{\fcolorbox{lime}{black}{\blue{AnsweR:-}}}$

Explanation:

given that,

Initial speed of the car ,u = 120 km/hr = 33.33 m/s

Finally it comes to rest, v = 0

Distance ,s = 100m

a) Let a is the average retardation, Using third equation of motion as:

$\begin{gathered}v {}^{2} - u {}^{2} = 2as \\ \\a = \frac{ { - u}^{2} }{2s} \\ \\ a = \frac{ { - (33.33)}^{2} }{2 \times 100} \\ \\ a = - 5.55m/s {}^{2} \end{gathered}$

b) Leti is the time taken by the car to come to rest. It is given by:

$\begin{gathered}t = \frac{v - u}{a} \\ \\ t = \frac{ - 33.33}{ - 5.55} \\ \\ t = 6 \: sec\end{gathered}$

Hence, this is the required equation

$\underline\pink{Hope\: it\: helps\:you}$

$\huge\colorbox{orange}{kнƲşнì❤࿐}$

${\mathbb{\colorbox {red} {\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {lime}{\boxed{\boxed{\boxed{\boxed{\boxed{\colorbox {aqua} {@XxItzSmartGirlxX}}}}}}}}}}}}}}}$

Answer 2

♡Answer:-♡࿐

A car moving along a straight road with a speed of 120 km/hr, is brought to rest by applying brakes, The car covers a distance of 100 m before it stops.

☆To find:-

. The average retardation of the car

. Time taken by the car to come to rest

Solution :-

Initial velocity (u) =120 km/h

final velocity (v) = 0

Distance covered (s) =100m

➞ u = 120 × 5/18

➞ u= 100 /3 m /s

According to third equation of motion.

➞ v2 = u2 + 2as

➞ (0) = (10000) /3) + 2×a×100

➞ 0 = 10000 /9 = 200a

➞ a = -10000/9×200

➞ a = - 1000/9 × 200

➞a = -100/18

➞a=-50/9

☆Hence,

Retardation of a car is 50/9m/s

First equation of motion

➞V=u+at

➞V =U + at

➞0=100/3+(-50) 19×t

➞0 = 100/3-50t

➞50t/9=100/3

➞t=9×100/50×3

➞t=3×2

➞t=6 sec

☆Therefore:-

Time taken =6 sec

Hope it help uh sweetu☺✌

I mich uh karan ╥﹏╥