Question

Q4. A car accelerates uniformly from 20 Km/h to 48 Km/h in 25 sec. Calculate i) The acceleration
ii) Distance cover by the car in that time.

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Answer 1

Answer :

➥ Acceleration of a car = 0.31 m/s²

➥ Distance cover by a car = 95.75 m

Given :

➤ Intial velocity of a car (u) = 20 km/h

➤ Final velocity of a car (v) = 48 km/h

➤ Time taken by a car (t) = 25 sec

To Find :

➤ Acceleration of a car (a) = ?

➤ Distance cover by a car (s) = ?

Solution :

◈ Intial velocity (u) = 20 km/h = 5.5 m/s

◈ Final velocity (v) = 48 km/h = 13.3 m/s

Acceleration of a car

Acceleration is given by

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

$\tt{: \implies a = \dfrac{13.3 - 5.5}{25} }$

$\tt{: \implies a = \cancel{\dfrac{7.8}{25} }}$

$\bf{: \implies \underline{ \: \: \underline{\purple{ \: \: a = 0.31 \: m/s^2 \: \: }} \: \: }}$

Hence, the acceleration of a car is 0.31 m/s².

Distance cover by a car

From second equation of motion

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

$\tt{: \implies s = 0 \times 25 + \dfrac{1}{ \cancel{ \: 2 \: }} \times 0.31 \times \cancel{25}\times 25}$

$\tt{: \implies s = 0 + 1 \times 0.31 \times 12.5 \times 25}$

$\tt{: \implies s = 0 + 0.31 \times 12.5 \times 25}$

$\tt{: \implies s = 0 + 3.87 \times 25}$

$\bf{: \implies \underline{ \: \: \underline{\purple{ \: \: s = 96.75 \: m\: \: }} \: \: }}$

Hence, the distance covered by car is 95.75 m.

⠀

Some releted equations :

⪼ s = ut + ½ at²

⪼ v = u + at

⪼ v² = u² + 2as

Answer 2

Answer:

Acceleration = 0.31m/s²

Distance = 95.75m