Question

A particle moving with constant acceleration covers a distance of 30m in 3rd sec and a distance of 50m in 5th sec what is the initial velocity

Answer 1

Given:

Distance travelled in 3rd sec= 30 m

Distance travelled in 5th sec= 50 m

To Find:

Initial velocity of particle

Solution:

We know that,

• Distance covered in nth second is given by

$\pink{\underline{\boxed{\bf{S_{n}=u+\frac{1}{2}a(2n-1) }}}}$

where,

u is initial velocity

a is acceleration

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According to question,

$\longrightarrow\rm{S_{3}=u+\dfrac{1}{2}a(2(3)-1)}$

$\longrightarrow\rm{30=u+\dfrac{1}{2}a(6-1)}$

$\longrightarrow\rm{30=u+\dfrac{5a}{2}}$

$\longrightarrow\rm{60=2u+5a}$

$\longrightarrow\rm{2u+5a=60}------(1)$

Also, according to question,

$\longrightarrow\rm{S_{5}=u+\dfrac{1}{2}a(2(5)-1)}$

$\longrightarrow\rm{50=u+\dfrac{1}{2}a(10-1)}$

$\longrightarrow\rm{50=u+\dfrac{9a}{2}}$

$\longrightarrow\rm{100=2u+9a}$

$\longrightarrow\rm{2u+9a=100}------(2)$

On subtracting (1) from (2), we get

$\longrightarrow\rm{4a=40}$

$\longrightarrow\rm{a=\dfrac{40}{4}}$

$\longrightarrow\rm{a=10\ m/s^{2}}$

On putting value of a in (1), we get

$\longrightarrow\rm{2u+5(10)=60}$

$\longrightarrow\rm{2u+50=60}$

$\longrightarrow\rm{2u=10}$

$\longrightarrow\rm\green{u=5\ m/s}$

Answer 2

Given:

• A particle moving with constant speed cover a distance of 30m in 3rd second and a distance of 50 m in 5th second:]

We know, that

• Acceleration=change in velocity/total time taken
• Velocity=change in distance/total time taken

Solution,

• As per the given clue in the question , we have to assume Distance ¹=30m Time¹=3sec Distance ²=50m Time ²=5 sec

Here,

• we can indicate distance in the form S and time for t]
• S¹=30m T¹=3sec S²=50m T²=5sec

Now calculate velocity here:]

• Velocity=Change in distance/total time taken
• velocity=S²-S¹/T²-T¹
• Velocity=50-30/5-3=>20/2=10m/s

Now calculate acceleration here:]

• Acceleration=Change in velocity/Total time taken
• Acceleration=10/2=>5m/s²

Hence,

• The acceleration of moving particle is 5m/s²