Question

and and object is started from rest with constant acceleration of 3m/s² for 10s. Then, find out final velocity after 10s also distance covered by object in 10s.​

Answer 1

Question:-

An object is started from rest with constant acceleration of 3m/s² for 10s. Then, find out final velocity after 10s also distance covered by object in 10s.

Answer:-

$\boxed{\bf{\color{h}{\: final \: velocity \: = \: 30m/s \: }}}$

$\boxed{\bf{\color{h}{\: distance \: = \: 150m \: }}}$

Solution:-

Given,

• An object starts from rest and accelerates of 3m/s² for 10sec.

We can find the final velocity by using the first equation of motion:-

u = 0 ; a = 3m/s² ; t = 10sec ; v = ?

Here,

• u = initial velocity
• v = final velocity
• a = acceleration
• s = distance
• t = time

➣ v = u + at

➣ v = 0 + 3(10)

$\boxed{\bf{\color{h}{\: v \: = \: 30m/s \: }}}$

Now,

We can find the distance by using the third equation of motion:-

➣ s = ut + 1/2 at²

➣ s = 0(10) + 1/2 × 3 × 10 × 10

➣ s = 3 × 5 × 10

$\boxed{\bf{\color{h}{\: s \: = \: 150m \: }}}$

Answer 2

Answer:-

$\red{\bigstar}$ Final velocity $\large\leadsto\boxed{\rm\green{30 \: m/s}}$

$\red{\bigstar}$ Distance covered $\large\leadsto\boxed{\rm\green{150 \: m}}$

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• Given:-

• Initial velocity = 0 m/s [starts from rest]

• Acceleration = 3 m/s²

• Time = 10 sec.

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• To Find:-

• Final velocity

• Distance covered

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• Solution:-

✯ Final velocity:-

We know,

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• First Equation of motion:-

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\purple{v = u + at}}}$

where,

• v = final velocity

• u = initial velocity

• a = acceleration

• t = time taken

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• Substituting in the values:-

➪ $\sf v = 0 + 3 \times 10$

➪ $\sf v = 0 + 30$

★ $\bf\red{v = 30 \: m/s}$

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Therefore, final velocity is 30 m/s.

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✯ Distance covered:-

We know,

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• Second Equation of motion:-

$\pink{\bigstar}$ $\large\underline{\boxed{\bf\purple{s = ut + \dfrac{1}{2}at^2}}}$

where,

• s = distance covered

• u = initial velocity

• t = time taken

• a = acceleration

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• Substituting in the values:-

➪ $\sf s = 0 \times 10 + \dfrac{1}{2} \times 3 \times (10)^2$

➪ $\sf s = 0 + \dfrac{1}{2} \times 3 \times 100$

➪ $\sf s = 0 + 3 \times 50$

★ $\bf\red{s = 150 \: m}$

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Therefore, the distance covered is 150 m.