Question

9. A car accelerates uniformly from 18 km 36 km/h in 2 seconds. Calculate a. the acceleration b. the distance covered by the car in that time. Pls give the full explanation..I will mark u as the brain liest​

Answer 1

Answer:

Explanation:

Solution,

• Initial velocity, u = 18 km/h = 18 × 5/18 = 5 m/s.
• Final velocity, v = 36 km/s = 36 × 5/18 = 10 m/s.
• Time taken, t = 2 seconds,

To Find,

• Acceleration, a and
• Distance covered. s

Formula to be used,

• 1st equation of motion, v = u + at
• 3ed equation of motion, v² - u² = 2as

Solution,

Putting all the values, we get

⇒ v = u + at

⇒ 10 = 5 + a × 2

⇒ 10 - 5 = 2a

⇒ 5 = 2a

⇒ 5/2 = a

⇒ a = 2.5 m/s².

Hence, the acceleration of the car is 2.5 m/s².

Now, we have to find the distance covered, s

Putting all the values again, we get

⇒ v² - u² = 2as

⇒ (10)² - (5)² = 2 × 2.5 × s

⇒ 100 - 25 = 5s

⇒ 75 = 5s

⇒ 75/5 = s

⇒ s = 15 m.

Hence, the distance covered by car is 15 m.

Answer 2

Answer:

Given :-

• A car accelerates uniformly from 18 km/h to 36 km/hr in 2 seconds.

To Find :-

• What is the acceleration.
• What is the distance covered by the car in that time.

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v =\: u + at}}}$

$\clubsuit$ 3rd Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2as}}}$

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken
• s = Distance Covered

Solution :-

First, we have to change initial and final velocity km/h into m/s :

❒ Initial Velocity (v) :

$\implies \sf Initial\: Velocity =\: 18\: km/h$

$\implies \sf Initial\: Velocity =\: {\cancel{18}} \times \dfrac{5}{\cancel{18}}\: m/s\: \: \bigg\lgroup \small\sf\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies\sf\bold{\purple{Initial\: Velocity =\: 5\: m/s}}$

❒ Final Velocity (u) :

$\implies \sf Final\: Velocity =\: 36\: km/h$

$\implies \sf Final\: Velocity =\: {\cancel{36}} \times \dfrac{5}{\cancel{18}}\: m/s\: \: \bigg\lgroup \small\bold{\pink{1\: km/h =\: \dfrac{5}{18}\: m/s}}\bigg\rgroup$

$\implies \sf Final\: Velocity =\: 2 \times 5\: m/s$

$\implies \sf\bold{\purple{Final\: Velocity =\: 10\: m/s}}$

Now, we have to find the acceleration :

Given :

• Final Velocity (v) = 10 m/s
• Initial Velocity (u) = 5 m/s
• Time (t) = 2 seconds

According to the question by using the formula we get,

$\longrightarrow \bf{v =\: u + at}$

$\longrightarrow \sf 10 =\: 5 + a(2)$

$\longrightarrow \sf 10 =\: 5 + 2a$

$\longrightarrow \sf 10 - 5 =\: 2a$

$\longrightarrow \sf 5 =\: 2a$

$\longrightarrow \sf \dfrac{5}{2} =\: a$

$\longrightarrow \sf 2.5 =\: a$

$\longrightarrow \sf\bold{\red{a =\: 2.5\: m/s^2}}$

${\small{\bold{\underline{\therefore\: The\: acceleration\: of\: a\: car\: is\: 2.5\: m/s^2\: .}}}}$

Now, we have to find the distance covered by the car in that time :

Given :

• Final Velocity (v) = 10 m/s
• Initial Velocity (u) = 5 m/s
• Acceleration (a) = 2.5 m/s²

According to the question by using the formula we get,

$\longrightarrow \bf v^2 =\: u^2 + 2as$

$\longrightarrow \sf (10)^2 =\: (5)^2 + 2(2.5)s$

$\longrightarrow \sf 10 \times 10 =\: 5 \times 5 + 2 \times \bigg(\dfrac{25}{10}\bigg) \times s$

$\longrightarrow \sf 100 =\: 25 + \dfrac{5\cancel{0}}{1\cancel{0}} \times s$

$\longrightarrow \sf 100 =\: 25 + 5s$

$\longrightarrow \sf 100 - 25 =\: 5s$

$\longrightarrow \sf 75 =\: 5s$

$\longrightarrow \sf \dfrac{\cancel{75}}{\cancel{5}} =\: s$

$\longrightarrow \sf \dfrac{15}{1} =\: s$

$\longrightarrow \sf 15 =\: s$

$\longrightarrow\sf\bold{\red{s =\: 15\: m}}$

${\small{\bold{\underline{\therefore\: The\: distance\: covered\: by\: the\: car\: in\: that\: time\: is\: 15\: m\: .}}}}$