Question

$\large{\underline{\underline{\red{\bf{Question}}}}}$
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→ A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s⁻² for 8.0 s. How far does the boat travel during this time ?​

Answer 1

Solution -

We have,

• Initial velocity (u) = 0
• Time (t) = 8.0 s
• Acceleration (a) = 3.0 ms⁻²

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In this question, we have to find the distance travelled by the boat during the given time.

We will use the second equation of motion, i.e.,

$\bf\longmapsto{\boxed{\pink{s = ut + \dfrac{1}{2}at^2}}}$

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Here,

• s = distance travelled
• u = initial velocity
• t = time taken
• a = acceleration

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Putting values in the formula

$\sf\dashrightarrow{s = 0 \times 8.0 + \dfrac{1}{2} \times 3 \times (8.0)^2}$

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$\sf\dashrightarrow{s = 0 + \dfrac{3}{2} \times 64}$

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$\sf\dashrightarrow{s = 3 \times 32}$

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$\frak\dashrightarrow{\purple{s = 96\:m}}$

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★ $\small\underline{\sf{Thus,\: the\: boat\: travels\: a\: distance\: of\: 96\: metres.}}$

Answer 2

Answer:

GIVEN:

Initial Velocity (u) = 0

Acceleration (a) = 3 m/s²

Time taken (t) = 8 seconds

TO FIND:

What is the distance covered by the boat in 8 seconds ?

SOLUTION:

Let the distance travelled by the boat be 's' m

To find the distance of the boat, we use the formula:-

$\large\bf{\star \: s = ut + \dfrac{1}{2} at^2 \: \star}$

According to question:-

$\sf{\rightarrow s = 0 \times 8 + \dfrac{1}{2} \times 3 \times (8)^2}$

$\sf{\rightarrow s = 0 + \dfrac{1}{\cancel{2}} \times 3 \times \cancel{8} \times 8}$

$\sf{\rightarrow s = 3 \times 4 \times 8}$

$\bf{\rightarrow \star \: s = 96 \: m \: \star}$

❝ Hence, the distance travelled by the boat is 96 m ❞

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✯ Extra Information ✯

➩ First equation of motion = v = u + at

➩ Third Equation of motion = v² - u² = 2as