Question

A car starts from rest travels with an uniform acceleration.The car travels 100 meter in 5 seconds.What will be it's final velocity?​

Answer 1

Given:-

• Initial velocity of car (u) = 0m/s.

• Distance traveled by car (s) = 100m

• Time taken = 5s.

To Find:-

• Final Velocity (v)

Solution:-

Firstly we find the acceleration of car .

So, By using 2nd equation of motion

⟹s = ut +1/2at².

Put the values ,we get

⟹ 100= 0×5 +1/2×a×5²

⟹ 100= 1/2×a× 25

⟹ 100 ×2 = a×25

⟹ a = 200/25

⟹ a = 8m/s²

∴ The acceleration of car is 8m/s²

And Now , by using 1st equation of motion

⟹ v = u+at

put the values,we get

⟹ v = 0+8×5

⟹v = 40m/s

∴ The Final velocity of the car is 40m/s.

Answer 2

AnswEr:

☯ Given:-

$\qquad\bullet\normalsize\sf\ Initial \: velocity \: of \: car = 0m/s$

$\qquad\bullet\normalsize\sf\ Distance \: travelled\: by \: car = 100m$

$\qquad\bullet\normalsize\sf\ Time\: taken \: by \: car = 5s$

☯ To Find:-

$\qquad\bullet\normalsize\sf\ Final \: velocity \: of \: car$

☯ Solution:-

$\underline{\bigstar\:\sf{According \: to \: given \: in \: question:}}$

First we need to find acceleration in order to find final velocity of car, so Let's begin with " second equation of motion";

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

Put the known values in it;

$\normalsize\dashrightarrow\sf\ 100 = 0 \times\ 5 +\frac{1}{2} \times\ a \times\ 5^{2}$

$\normalsize\dashrightarrow\sf\ 100 = 0 +\frac{1}{2} \times\ a \times\ 25$

$\normalsize\dashrightarrow\sf\frac{100 \times\ 2}{25} = a$

$\normalsize\dashrightarrow\sf\frac{\cancel{200}}{\cancel{25}} = a$

$\normalsize\dashrightarrow\sf\ 4 = a$

$\normalsize\dashrightarrow\sf\ a = 4m \setminus\ s^{2}$

Now, put the known values and value of acceleration in "first equation of motion";

$\normalsize\dashrightarrow\sf\ v = u + at$

$\normalsize\dashrightarrow\sf\ v = 0 + 4 \times\ 5$

$\normalsize\dashrightarrow\sf\ v = 0 + 20$

$\normalsize\dashrightarrow\sf\ v = 20m/s$

$\therefore\:\underline{\textsf{Hence, \: the \: final \: velocity \: is}{\textbf{\: 20m/s}}}$