Question

- A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 30m
${s}^{ - 2}$
for 8.0s. How far does the boat travel during this time? ​

Answer 1

Given :-

Initial velocity (u) = 0 [ It started from rest ]

Acceleration (a)= 30m/s^2

Time (t) = 8 seconds

To find :-

We have to find the distance covered by boat in the given time

$\underline{\bigstar{\sf{By\ using\ 2nd \ equation\ of\ motion}}}$

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

where ,

u= Initial velocity

a= acceleration

t= time

s= distance

• Put the given values !

$:\implies\sf s= ut+\dfrac{1}{2}at^2\\ \\ :\implies\sf s= 0\times (8)+ \dfrac{1}{\cancel{2}}\times \cancel{30}\times (8)^2\\ \\ :\implies\sf s= 0+ 15\times 64\\ \\ :\implies\sf s= 960m$

$\underline{\boxed{\textsf{\textbf{\ \ Distance \ covered \ by \ boat = 960m}}}}$

Answer 2

The boat travels 960 m during that time.

Solution

We have motorboat starting from rest accelerating at rate of 30 m/s² for 2 seconds.

We have to find distance covered by motorboat during that time.

Here, u = 0 m/s [∵It stars from rest]

Acceleration (a) = 30 m/s²

Time (t) = 8 seconds.

☢ According to 1st equation of motion :

● v = u + at

⇒ v = 0 + 30 × 8

⇒ v = 0 + 240

⇒ v = 240 m/s

∴ Final velocity of boat = 240 m/s

☢ Now using 3rd equation of motion :

● v² - u² = 2as

⇒ 240² - 0² = 2 × 30 × s

⇒ 57600 = 60s

⇒ s = 57600/60

⇒ s = 960 m

∴ Distance covered by boat during this time = 960 m.