Question

A motorboat starting from west on a lake accelerates in a straight line at a constant rate of 4 m/s2 for 9 seconds. How far does the boat travel during this time?​

Answer 1

Answer:

s = 1/2at^2

=1/2×4×9×9

=81×2

=162

Answer 2

Answer:

• Distance travelled is 162 m.

Given:

• Acceleration = 4 m/s²
• Time taken = 9 sec

Explanation:

$\rule{300}{1.5}$

Here, we are given that a motorboat starts from rest on a lake accelerates in a straight line at a constant rate of 4 m/s² for 9 sec. We are asked to find the distance travelled by it.

Since, it starts from rest, so it's initial velocity will be zero. Firstly, let us find out the final velocity of the boat.

From first kinematic equation we know,

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

Substitute the values,

$\longrightarrow{\sf{v\ =\ 0\ +\ 4 \times 9}} \\ \\ \\ \longrightarrow{\sf{v\ =\ 0\ +\ 36}} \\ \\ \\ \longrightarrow{\sf{v\ =\ 36\ m/s}}$

∴ The final velocity of the boat is 36 m/s.

$\rule{300}{1.5}$

Now, let us find out the distance travelled by the boat.

From third kinematic equation we know,

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

Substitute the values,

$\longrightarrow{\sf{(36)^2\ =\ (0)^2\ +\ 2 \times 4 \times s}} \\ \\ \\ \longrightarrow{\sf{1,296\ =\ 0\ +\ 8s}} \\ \\ \\ \longrightarrow{\sf{\dfrac{1,296}{8}\ =\ s}} \\ \\ \\ \longrightarrow{\sf{162\ =\ s}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 162\ m}}$

∴ The distance travelled by the boat is 162 m.

$\rule{300}{1.5}$

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

From second kinematic equation we know,

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

Solving,

$\longrightarrow{\sf{s\ =\ 0 \times 9\ +\ \dfrac{1}{2} \times 4 \times (9)^2}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 0\ +\ \dfrac{1}{2} \times 4 \times 9 \times 9}} \\ \\ \\ \longrightarrow{\sf{s\ =\ \dfrac{1}{2} \times 4 \times 81}} \\ \\ \\ \longrightarrow{\sf{s\ =\ \dfrac{1}{\cancel2} \times \cancel4 \times 81}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 2 \times 81}} \\ \\ \\ \longrightarrow{\sf{s\ =\ 162\ m}}$

∴ The distance travelled by the boat is 162 m.

$\rule{300}{1.5}$