Question

A car accelerates uniformly from 18km/hr to 16km/hr in 5s.calculate the acceleration and distance covered in that time

Answer 1

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HERE'S THE ANSWER...

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♠️ We'll note the given data.

⏺️ Initial velocity ( u ) = 18 Km / h

=> 18 × 5 / 18 = 5 m / s

⏺️Final velocity ( v ) = 16 Km / h

=> 16 × 5 / 18 = 4.44 m / s

⏺️ Time ( t ) = 5 second

⏺️ Acceleration ( a ) = ?

♠️ We know Acceleration is know as change in velocity with respect to time , i.e

✔️ Acceleration = ( Final velocity - initial velocity ) / time

=> a = ( v - u ) / t

=> a = ( 4.44 - 5 ) / 5

=> $\boxed{a\:=\:-0.11\: m/s^2}$

⏺️the vehicle is retarding

▶️ And we'll find distance in 5 second with eqn

✔️ ( v )^2 - ( u )^2 = 2 × a × s

=> ( 4.44 )^2 - ( 5 )^2 = 2 × ( - 0.11 ) × s

=> 19.71 - 25 = - 0.22 × s

=> -5.28 / - 0.22 = s

=> $\boxed{s\:=\:24.02\: m}$

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Answer 2

Given,

u = 18 km/hr ( 18 x 5/18 m/s = 5 m/s)

=> 5 m/s

v = 16 km/hr (16 x 5/18 = 40/9)

 => 40/9 m/s

so, acceleration

=> a = (v - u) / t

=> a = 40/9 - 5 /5

=> a = -5/9 /5

=> a => -1/9

Note: Here minus sign shows that vechile's speed is retarding.

Now,

the total distance travelled will be

s = ut + (1/2)at2

or

=> s = 5x5 + (1/2).(-1 /9).(5)2

=> s = 25+(-1/2 x 1/9 x 25)

=> s = 25 + (-25/18)

=> s = 25 x 18 - 25/18

=> s = 25-25/18

=> s = 450-25/18

=> s = 425/18

=> s = 23.61

or

about 23.62 m

thus,

s = 23.62m

Therefore,

Distance covered in 5 seconds is about 23.62 metres.

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