Physics, asked by Zonariah4838, 8 months ago

A car accelerates uniformly from 18 km h-¹ to 36 I'm h-¹ in 5s. Calculate:- (I) the Acceleration (II) Distance covered by the car in that time. PLEASE ANSWER, IT IS VERY URGENT

Answers

Answered by Anonymous
1086

Given,

 \tt u = 18   \:  km \   {h}^{ - 1 } = 5 \: m \:  {s}^{ - 1}

 \tt v = 36  \:  km \   {h}^{ - 1 } = 10 \: m \:  {s}^{ - 1}

 \tt t  = 5 \: s

(I) Calculating Acceleration

As, we know

 \huge \boxed {  \boxed{\tt { a =  \frac{v - u}{t}} }}

 \tt \implies  \dfrac{10 \: m \:  {s}^{ - 1 }  \:  -  \: 5 \: m \:  {s}^{ - 1} }{5 \: s}

 \tt \implies \dfrac{5 \: m \:  {s}^{ - 1} }{5 \: s}

 \tt \implies 1 \: m \:  {s}^{ - 1}

Thus, acceleration is  \rm 1 \: m \:  {s}^{ - 1}

(II) Calculating Distance Travelled

As, we know

  \huge \boxed {  \boxed{\tt { s =ut +  \frac{1}{2}   \: at ^{2}  }}}

 \tt \implies 5 \: m \:  {s}^{ - 1}  \times 5 \: s  \: +  \:  \dfrac{1}{2}  \times 1 \: m \:  {s}^{ - 2}  \times (5 \: s) ^{2}

 \tt \implies 25 \: m \:   +  \:  \dfrac{25}{2}  \: m

 \tt \implies 25 \: m \:  +  \: 12.5 \: m

 \tt \implies 37.5 \: m

Thus, distance travelled is   \rm 37.5 \: m

Additional Information

  • Rate of Change of Distance is known as Speed.

  • The speed of body which is applied in beginning is known as Initial Speed and is denoted by u.

  • The speed of body which is acquired after the body starts Moving is know as Final Speed and is represented by v.

  • Rate of change of Velocity is known as Acceleration.

  • Negative Acceleration is known as Retardation.

Important Formulas

 \bf Speed = \dfrac{Distance}{Time}

 \bf Average \: Speed = \dfrac{Total \: distance}{Total \: time}

 \bf v = u + at

 \bf v^{2} - u^{2} = 2as

 \bf S = ut + 1/2at^{2}


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Answered by Anonymous
109

Explanation:

Acceleration=change in velocity/ time

Change in velocity=36-18=18km/h=5m/s

time=5s

acceleration=5/5=1m/s2

equation of motion,

s=ut+(1/2)at2

u=18 km/h=5 m/s

t=5s

a=1m/s2

s=(5*5)+(1/2*1*5*5)

s=25+12.5 i.e.,=37.5m

formula used:

change in velocity\time=acceleration

s=ut+(1/2) at2

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