Question

Body starting with initial velocity of 5 metre per second moving with a uniform acceleration of 2 metre per second square after some time and attains a velocity of 30 m per square find the distance covered by the body during the period

Answer 1

Given

• Initial velocity (u) = 5m/s
• Acceleration (a) = 2m/s²
• Final velocity (v) = 30m/s

Find out

• Find the distance covered by body

Solution

$\star{\rm{\underline{\red{According\:to\:the\:third\:equation\:of\:motion}}}}$

➙ v² = u² + 2as

➙ (30)² = (5)² + 2 × 2 × s

➙ 900 = 25 + 4s

➙ 900 - 25 = 4s

➙ 875 = 4s

➙ s = 875/4

➙ s = 218.75 m

${\blue{\boxed{\sf{Distance\: covered\:by\:body=218.75}}}}$

Additional Information

• v = u + at {1st equation of motion}

• s = ut + ½ at² {2nd equation of motion}

Answer 2

GIVEN:

• Initial Velocity (u) of the body = 5 m/s
• Final velocity (v) of the body = 30 m/s
• Acceleration (a) = 2 m/s²

TO FIND:

• What is the distance covered by the body ?

SOLUTION:

Let the distance covered by the body be 's' m

To find the distance covered by the body, we use the third equation of motion:-

$\large\bf{\star \: v^2 - u^2 = 2as\: \star}$

According to question:-

On putting the given values in the equation, we get

$\rm{\dashrightarrow (30)^2 - (5)^2 = 2 \times 2 \times s }$

$\rm{\dashrightarrow 900 - 25 = 4s }$

$\rm{\dashrightarrow 875 = 4s }$

$\rm{\dashrightarrow \cancel\dfrac{875}{4} = s }$

$\bf{\dashrightarrow 218.75 \: m = s}$

❝ Hence, the distance covered by the body is 218.75 m ❞

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✰ Extra Information ✰

➺ First equation of motion = v = u + at

➺ Second equation of motion = s = ut + 1/2 at²