Question

auto speed from 20m/s to 50m/s with acceleration of 10m/s calculate the distance travelled and time taken?​

Answer 1

$\red{\underline{{ \bf Correct \: Question }}}$

• An auto speeds from 20m/s to 50m/s with acceleration of 10m/s². Calculate the distance travelled and time taken?

$\red{\underline{{ \bf Solution }}}$

$\underline {\underline{{\purple{ \sf Given }}}}$

• Initial Velocity (u) ➠ 20 m/s
• Final Velocity (v) ➠ 50 m/s
• Acceleration (a) ➠ 10 m/s²

$\underline {\underline{{ \purple{\sf To \: Find }}}}$

• Time Taken (t)
• Distance Travelled (s)

$\underline {\underline{{ \green{\sf Calculating \: Time \: Taken }}}}$

Formula Used :- $\underline{\underline{\boxed{ \gray{\sf v = u +at }}}}$

Substituting Values

☞ 50 = 20 + 10 × t

☞ 50 - 20 = 10 t

☞ 30 = 10 t

☞ $\sf t = \cancel{\dfrac{30}{10}}$

☞ t = 3

${\green{ \underbrace{\boxed{\underline{\underline{\purple{ \sf \therefore Time \: Taken \: is \: 3 \: sec}}}}}}}$

Now,

$\underline {\underline{{ \green{\sf Calculating \: Distance \: Travelled }}}}$

Method 1

Formula Used :- $\underline{\underline{\boxed{ \gray{\sf{v}^{2} - {u}^{2} = 2as }}}}$

Substituting Values

☞ (50)² - (20)² = 2 × 10 × s

☞ 2500 - 400 = 20s

☞ 2100 = 20s

☞ $\sf s = \dfrac{2100}{20}$

☞ 105

Method 2

Formula Used :- $\underline{\underline{\boxed{ \gray{\sf s = ut + \dfrac{1}{2} at^{2} }}}}$

Substituting Values

☞ s = 20 × 3 + ½ × 10 × 3²

☞ s = 60 + ½ × 10 × 9

☞ s = 60 + 5 × 9

☞ s = 60 + 45

☞ s = 105

${\pink{ \underbrace{\boxed{\underline{\underline{\orange{ \sf \therefore Distance \: Travelled \: is \: 105\: m}}}}}}}$

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Answer 2

Question :

An auto speeds from 20m/s to 50m/s with acceleration of 10m/s². Calculate the distance travelled and time taken?

To Find :

• Time Taken

• Distance Covered .

Given :

• Final velocity = 50 m/s

• Initial velocity = 20 m/s

• Acceleration = 10 m/s

We Know :

First EQuation of Motion :

$\blue{\sf{\underline{v = u + at}}}$

Where,

• v = Final velocity

• u = Initial velocity

• a = Acceleration due to gravity

• t = Time Taken

Second Equation of Motion :

$\blue{\sf{\underline{s = ut + \dfrac{1}{2}at^{2}}}}$

where,

• v = Final velocity
• u = Initial velocity

• a = Acceleration due to gravity

• t = Time Taken

• s = Distance Traveled

Third EQuation of Motion :

$\blue{\sf{\underline{v^{2} = u^{2} + 2as}}}$

• v = Final velocity

• u = Initial velocity

• a = Acceleration due to gravity

• s = Distance covered

Solution :

"Time Taken" :-

• v = 50 m/s
• u = 20 m/s
• a = 10 m/s

Using the first Equation of Motion , and by substituting the values in it , we get :

$\blue{\sf{v = u + at}} \\ \\ \\ \implies \sf{50 = 20 + 10 \times t} \\ \\ \\ \implies \sf{50 - 20 = 10 \times t} \\ \\ \\ \implies \sf{30 = 10 \times t} \\ \\ \\ \implies \sf{\dfrac{30}{10} = t} \\ \\ \\ \implies \sf{\dfrac{3\not{0}}{1\not{0}} = t} \\ \\ \\ \implies \sf{3 s = t} \\ \\ \\ \therefore \purple{\sf{t = 3 s}}$

Hence , the time taken is 3 seconds.

"Distance Covered" :-

• v = 50 m/s
• u = 20 m/s
• a = 10 m/s²

Using the Third Equation of Motion , and substituting the values in it , we get :

$\blue{\sf{v^{2} = u^{2} + 2as}} \\ \\ \\ \implies \sf{50^{2} = 20^{2} + 2 \times 10 \times s} \\ \\ \\ \implies \sf{50^{2} - 20^{2} = 20 \times s} \\ \\ \\ \implies \sf{2500 - 400 = 20 \times s} \\ \\ \\ \implies \sf{2100 = 20 \times s} \\ \\ \\ \implies \sf{\dfrac{2100}{20} = s} \\ \\ \\ \implies \sf{105 m = s} \\ \\ \\ \therefore \purple{\sf{s = 105 m}}$

Hence , the Distance Traveled is 105 m.

Alternative Method :-

"Distance Covered" :

• t = 3 s
• u = 20 m/s
• a = 10 m/s²

$\blue{\sf{s = ut + \dfrac{1}{2}at^{2}}} \\ \\ \\ \implies \sf{s = 20 \times 3 + \dfrac{1}{2} \times 10 \times 3^{2}} \\ \\ \\ \implies \sf{s = 60 + \dfrac{1}{2} \times 10 \times 9} \\ \\ \\ \implies \sf{s = 60 + 5 \times 9} \\ \\ \\ \implies \sf{s = 60 + 45} \\ \\ \\ \implies \sf{s = 105 m} \\ \\ \\ \therefore \purple{\sf{s = 105 m}}$

Hence , the Distance Covered is 105 m.