Question

a motorboat staring from rest on a lake accelerates in a straight line at a constant rate of 4metre par second^2for 6second .how far does the boat travel during this time?​

Answer 1

Answer:

u = 0 m/s

a = 4 m/s

t = 6 s

By Second Equation of Motion : s = ut + 1/2at²

s = 0×6 + 1/2 × 4 × (6)²

s = 0 + 2 × 36

s = 72 m

Answer 2

Answer:

• The distance travelled by the boat is 72 m.

Given:

• Acceleration = 4 m/s²
• Time = 6 sec

Explanation:

$\rule{300}{1.5}$

We are give that a motor boat starts from rest on a lake accelerates with a constant rate of 4 m/s² in a straight line for 6 sec. We are asked to find the distance travelled during the given time period.

From second kinematic equation we know,

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

Here,

• s denotes distance travelled
• u denotes initial velocity
• t denotes time
• a denotes acceleration

Since, motor boat is starting from rest. So the initial velocity of the boat will be zero.

Solving,

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

Substitute the values,

$\longrightarrow{\sf{s\ =\ 0 \times 6\ +\ \dfrac{1}{2} \times 4 \times (6)^2}} \\ \\ \\ \longrightarrow{\sf{s\ =\ \dfrac{1}{2} \times 4 \times 6 \times 6}} \\ \\ \\ \longrightarrow{\sf{s\ =\ \dfrac{1}{2} \times 4 \times 36}} \\ \\ \\ \longrightarrow{\sf{s\ =\ \dfrac{1}{\cancel2} \times \cancel4 \times 36}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 2 \times 36}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 72\ m}}$

∴ The distance travelled by the boat is 72 m.

$\rule{300}{1.5}$

A L T E R N A T E - M E T H O D

Here, firstly we will find the final velocity of the boat and then find the distance travelled by it.

From first kinematic equation we know,

${\longrightarrow{\sf{v\ =\ u\ +\ at}}}$

Here,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

Substitute the values,

$\longrightarrow{\sf{v\ =\ u\ +\ at}} \\ \\ \\ \longrightarrow{\sf{v\ =\ 0\ +\ 4 \times 6}} \\ \\ \\ \longrightarrow{\sf{v\ =\ 0\ +\ 24}} \\ \\ \\ \longrightarrow{\sf{v\ =\ 24\ m/s}}$

∴ We got the final velocity of the boat.

$\rule{300}{1.5}$

From third kinematic equation we know,

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

Here,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance travelled

Since, we know the final velocity of the boat. So let's apply it in the above equation.

Substitute the values,

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

Solving,

$\longrightarrow{\sf{(24)^2\ =\ (0)^2\ +\ 2 \times 4 \times s}} \\ \\ \\ \longrightarrow{\sf{576\ =\ 0\ +\ 8s}} \\ \\ \\ \longrightarrow{\sf{\dfrac{576}{8}\ =\ s}} \\ \\ \\ \longrightarrow{\sf{72\ =\ s}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 72\ m}}$

∴ The distance travelled by the boat is 72 m.

$\rule{300}{1.5}$