Question

a moterboat starts from rest and attains a velocity of 15 m/s in 5 sec. calculate the accelaration​

Answer 1

Explanation:

Here's the answer

Given =>

Initial velocity = 0 m/s

Final velocity = 15 m/s

Time taken = 5 second

To find =>

Acceleration = ?

Distance travelled = ?

We know that,

{\boxed {\boxed { \boxed { \sf{Acceleration = \frac{v - u}{t}} }}}}

Acceleration=

t

v−u

\sf{= \frac{15 - 0}{5}}=

5

15−0

\sf {= 3 \: m/s {}^{2}}=3m/s

2

Now,

By using second kinematical equation,

\sf{s = ut + \frac{1}{2}at {}^{2}}s=ut+

2

1

at

2

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

2

1

×3×25

\sf{= \frac{75}{2}}=

2

75

\sf{ = 37.5 \: m}=37.5m

Distance covered = 37.5 metre

Answer 2

Answer:

Initial velocity=0

Final velocity=15m/s

Time =5seconds

So,

\begin{gathered}acceleration = \frac{v - u}{t} \\ = > \frac{15 - 0}{5} \\ = > \frac{15}{5} \\ = > 3\end{gathered}acceleration=tv−u=>515−0=>515=>3

Acceleration=3m/s^2

Now,

we have to find Distance travelled,we will find it by using following equation:-

\begin{gathered}= > s = ut + \frac{1}{2} a {t}^{2} \\ = > s = 0 \times 5 + \frac{1}{2} \times 3 \times5 \times 5 \\ = > s = 0 + \frac{1}{2} \times 75\end{gathered}=>s=ut+21at2=>s=0×5+21×3×5×5=>s=0+21×75