Question

A motorboat starts from rest and moves with uniform acceleration. If it attains the

velocity of 5 m/s 5 s, calculate the acceleration and the distance travelled in that

time​

Answer 1

${\large{\pmb{\sf{\underline{Required \: solution...}}}}}$

${\bigstar \:{\pmb{\sf{\underline{About \: question...}}}}}$

⋆ Correct question: A motorboat starts from rest and moves with uniform acceleration. If it attains the velocity of 5 m/s 5 s, calculate the acceleration and the distance travelled in that time.

⋆ Understanding the question: This question says that we have to find out the acceleration and the distance travelled if a motorboat starts from rest and moves with uniform acceleration. And this motorboat attains the velocity of 5 m/s in 5 seconds. Let us solve this question!

${\bigstar \:{\pmb{\sf{\underline{Given \: that...}}}}}$

⋆ Final velocity of motorboat = 5 m/s

⋆ Initial velocity of motorboat = 0 m/s (As it is starts from rest then move)

⋆ Time taken = 5 seconds

${\bigstar \:{\pmb{\sf{\underline{To \: find...}}}}}$

⋆ The acceleration in that time

⋆ The distance travelled in that time

${\bigstar \:{\pmb{\sf{\underline{Solution...}}}}}$

⋆ The acceleration in that time = 1m/s sq.

⋆ The distance travelled in that time = 12.5 m

${\bigstar \:{\pmb{\sf{\underline{Using \: concept...}}}}}$

⋆ Formula to find acceleration

⋆ Second equation of motion

${\bigstar \:{\pmb{\sf{\underline{Using \: formula...}}}}}$

${\small{\underline{\boxed{\sf{\star \: a \: = \dfrac{(v-u)}{t}}}}}}$

${\small{\underline{\boxed{\sf{\star \: s \: = ut + \dfrac{1}{2} at^{2}}}}}}$

${\bigstar \:{\pmb{\sf{\underline{Where...}}}}}$

⋆ v denotes final velocity

⋆ u denotes initial velocity

⋆ t denotes time taken

⋆ s denotes displacement or distance

⋆ a denotes acceleration

${\bigstar \:{\pmb{\sf{\underline{Full \: Solution...}}}}}$

~ Firstly let us find the acceleration by using it's formula, we just have to put the values according to that formula and have to solve. Let's do it!

$:\implies \sf a \: = \dfrac{(v-u)}{t} \\ \\ :\implies \sf a \: = \dfrac{(5-0)}{5} \\ \\ :\implies \sf a \: = \dfrac{(5)}{5} \\ \\ :\implies \sf a \: = \dfrac{5}{5} \\ \\ :\implies \sf a \: = 1 \: m/s^{2}$

Hence, we get 1 m/s sq. as acceleration

~ Now let's find out the distance by using second equation of motion. We just have to put values according to it then we have to solve. Let's do it too!

$:\implies \sf s \: = ut + \dfrac{1}{2} at^{2} \\ \\ :\implies \sf s \: = 0(5) + \dfrac{1}{2} 1(5)^{2} \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} 1(25) \\ \\ :\implies \sf s \: = 0 + \dfrac{1}{2} 25 \\ \\ :\implies \sf s \: = \dfrac{1}{2} 25 \\ \\ :\implies \sf s \: = 12.5 \: m$

We get distance as 12.5 m

${\bigstar \:{\pmb{\sf{\underline{Explore \: more...}}}}}$

Distance = It is the length of actual path covered by a moving object in a given time interval.

Displacement = Shortest distance covered by a body in a definite direction is called displacement.

→ Displacement may be positive, negative or zero whereas distance is always positive.

→ Distance is a scaler quantity whereas displacement is a vector quantity both having the same unit.

→ In general magnitude of displacement ≤ Distance.

Speed = Distance travelled by a moving object in unit time interval is called speed i.e., speed Distance/Time. It's scaler quantity and it's SI unit is metre/second

Velocity = Velocity of moving object is defined as the displacement of the object in unit time interval i.e., velocity = Displacement/Time. It's vector quantity and it's SI unit is metre/second.

Acceleration = Acceleration of an object is defined as the rate of change of velocity of the object i.e., Acceleration = Change in velocity/Time. It's vector quantity and SI unit is m/s²

Answer 2

$\sf Given$

Initial velocity = 0 m/s

Final velocity = 15 m/s

Time taken = 5 second

• To find =>

Acceleration = ?

Distance travelled = ?

• We know that,

$\color{green}⟹ {\boxed {\boxed { \boxed { \sf{Acceleration= \frac{v - u}{t}} }}}}$

$➠\sf{= \frac{15 - 0}{5}}$

$\color{purple}➠\sf {= 3 \: m/s {}^{2}}=3m/s$

Now,

• By using second kinematical equation,

$➠\sf{s = ut + \frac{1}{2}at {}^{2}}$

$⟹ \sf{s = 0 + \frac{1}{2} \times 3 \times 25}$

$➠\sf{= \frac{75}{2}}$

$→\sf{ = 37.5 \: m}=37.5m$

Distance covered = 37.5 metre