Question

A car is moving with a velocity of 72 km/h .its velocity is reduced to 36 km/h after covering a distance of200m .calculate retardation .

Answer is 75 cm/s^2

Answer 1

Answer:

here the answer is in negative because it is retardation

and it is 3/4 m/s^2

or 75cm/s^2

ok

Explanation:

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guys

Answer 2

Answer:

$\displaystyle \implies a=75 \ cm/sec^2$

Explanation:

Given :

Initial velocity ( u ) = 72 km / hr.

To convert km / hr multiply it by 5 / 18 we get

u = 4 × 5 = 20 m / sec

Final velocity ( v ) = 36 km / hr

In terms of m / sec

v = 2 × 5 = 10 m / sec

Distance ( s ) = 200 m

From third equation of motion we have

$\displaystyle v^2=u^2+2as$

u = Initial velocity  

a = Acceleration

v = final velocity  

s = distance

put the value here we get

$\displaystyle 10^2=20^2+2\times200\times a\\\\\displaystyle 100=400+400\times a\\\\\displaystyle 400a=100-400\\\\\displaystyle 400a=-300\\\\\displaystyle a=-\frac{300}{400} \ m/sec^2\\\\\displaystyle a=-\frac{3}{4} \ m/sec^2$

Negative acceleration  means acceleration  in opposite direction also retardation .

To convert meter into centimeter.

$\displaystyle a=\frac{3}{4} \ m/sec^2\\\\\displaystyle \implies a=\frac{3}{4}\times100 \ cm/sec^2\\\\\displaystyle \implies a=3\times25 \ cm/sec^2\\\\\displaystyle \implies a=75 \ cm/sec^2$

Thus retardation of car is 75 $cm/sec^2$.