Question

A car accelerates uniformly from 18km to 36km per hour in 5 sec. Calculate (1) The acceleration and (2) distance corvered by car in that time​

Answer 1

Explanation:

a=v-u/t, a=36-18/5=18/5km

a=18/5km

distance=18+36=54km

Answer 2

Answer :

➥ The acceleration of a car = 1 m/s²

➥ The distance covered by a car = 37.5 m

Given :

➤ Intial velocity of a car (u) = 18 km/h

➤ Final velocity of a car (v) = 36 km/h

➤ Time taken by a car (t) = 5 sec

To Find :

➤ Acceleration of a car (a) = ?

➤ Distance covered by a car (s) = ?

Solution :

◈ Intial velocity (u) = 18 × 5/18 = 5 m/s

◈ Final velocity (v) = 36 × 5/18 = 10 m/s

Acceleration of a car

Acceleration is given by

$\tt{: \implies a = \dfrac{v - u}{t} }$

$\tt{: \implies a = \dfrac{10 - 5}{5} }$

$\tt{: \implies a = \cancel{\dfrac{5}{5}}}$

$\bf{: \implies \underline{ \: \: \underline{ \purple{ \: \: a = 1 \: m/s^2 \: \: }} \: \: }}$

Hence, the acceleration of a car is 1 m/s².

Distance corvered by car in that time

From second equation of motion

$\tt{: \implies s = ut + \dfrac{1}{2}a {t}^{2} }$

$\tt{: \implies s = 5 \times 5 + \dfrac{1}{2} \times 1 \times 5 \times 5}$

$\tt{: \implies s = 25 + \dfrac{1}{ \cancel {\:2 \: }} \times 1 \times \cancel{25 }}$

$\tt{: \implies s = 25 + 1 \times 12.5}$

$\tt{: \implies s = 25 + 12.5}$

$\bf{: \implies \underline{ \: \: \underline{ \purple{ \: \: s = 37.5 \: m \: \: }} \: \: }}$

Hence, the distance covered by a car is 37.5 m.

⠀⠀⠀⠀

Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as