Question

a car accelerate uniformly from 18km/h to 36km/h in 5s calculate 1 acceleration distance covered by car

Answer 1

For calculating acceleration, speed should be in m/s
Hence 18km/hr
=
$18 \times \frac{5}{18} \\ = 5$
5m/s

36km/hr =
$36 \times \frac{5}{18} = 10$
10 m/s

We know that acceleration =
$\frac{v - u}{t}$
where v is final velocity
u is initial velocity
And t is time

Hence acceleration =
$\frac{10 - 5}{5} \\ = \frac{5}{5} \\ = 1$
so acceleration = 1m/s²


Now distance travelled during acceleration (s) =
$ut + \frac{1}{2} at {}^{2}$
So this means
$5 \times 5 + \frac{1}{2} \times 1 \times 25 \\ = 25 + 12.5 \\ = 37.5$
So Distance travelled is 37.5m

Hope it helps dear friend ☺️✌️✌️

Answer 2

$\huge\mathfrak\pink{Solution}$

★ We know that

${ \ U \ = \ 18 \ km h¯¹ \ = \ 5 m s¯¹}$

${ \ v \ = \ 36 km h¯¹ \ = \ 10m s¯¹}$

${ \ t \ = \ 5s}$

${ \ 1)}$ ${ \ we \ have}$

${ \ a \ =}$ $\frac{v - u}{t}$

${ \ =}$ $\frac{10m s¯¹ - 5m s¯¹}{5s}$

${ \ 1m s¯¹}$

${ \ 2)}$ ${ \ we \ have}$

${ \ s \ = \ UT \ +}$ $\frac{1}{2}$ at²

${ \ 5m s¯¹ \ × 5s \ + \ +}$ $\frac{1}{2}$ ${ \ × \ 1m s¯¹ \ × \ (5 s)²}$

$\tt{\implies}$ ${ \ 25m \ + \ 12.5m}$

$\tt{\implies}$ ${ \ 37.5m}$

The acceleration of the car is 1m s¯² and the distance covered is 37.5 m

$\huge\mathfrak\pink{hope \ it's \ helps \ you}$