Question

A car accelerate uniformly from 18 km/h to 36 km/h in 5sec calculate.(1)the acceleration and (2)the distance covered by the car in that time.​

Answer 1

Given that:

• A car accelerate uniformly from 18 km/h to 36 km/h in 5 sec.

To Find:

1. The acceleration.
2. The distance covered by the car in that time.

We know that:

• a = (v - u)/t
• s = ut + (½)at²
• [ 1 km/hr = 5/18 m/sec ]

Where,

• a = Acceleration
• v = Final velocity
• u = Initial velocity
• t = Time
• s = Displacement

We have:

• v = 36 km/hr = (36 × 5)/18 = 10 m/sec
• u = 18 km/hr = (18 × 5)/18 = 5 m/sec
• t = 5 sec

Finding the acceleration:

⟶ a = (10 - 5)/5

⟶ a = 5/5

⟶ a = 1

Hence,

• Acceleration = 1 m/s²

Finding the distance covered by the car:

⟶ s = (5 × 5) + (½) × 1 × (5)²

⟶ s = 25 + (½) × 1 × 25

⟶ s = 25 + (½) × 25

⟶ s = 25 + 12.5

⟶ s = 37.5

Hence,

• The distance covered by the car is 37.5 m.

Answer 2

Answer:

Given :-

• A car accelerate uniformly from 18 km/h to 36 km/h in 5 seconds.

To Find :-

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

Formula Used :-

$\clubsuit$ First Equation Of Motion Formula :

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

$\clubsuit$ Second Equation Of Motion Formula :

$\mapsto \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2} at^2}}}$

where,

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

Solution :-

First, we have to convert km/h into m/s :

${\normalsize{\bold{\purple{\underline{\bigstar\: Initial Velocity\: (u)\: :-}}}}}$

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

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

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

${\normalsize{\bold{\purple{\underline{\bigstar\: Final\: Velocity\: (v)\: :-}}}}}$

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

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

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

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

Now, we have to find the acceleration :

Given :

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

According to the question by using the formula we get,

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

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

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

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

$\longrightarrow \sf \dfrac{1}{1} =\: a$

$\longrightarrow \sf 1 =\: a$

$\longrightarrow \sf\bold{\red{a =\: 1\: m/s^2}}$

$\therefore$ The acceleration of a car is 1 m/s².

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

Given :

• Initial Velocity (u) = 5 m/s
• Time (t) = 5 seconds
• Acceleration (a) = 1 m/s²

According to the question by using the formula we get,

$\longrightarrow \sf s =\: (5)(5) + \dfrac{1}{2} \times (1)(5)^2$

$\longrightarrow \sf s =\: 5 \times 5 + \dfrac{1}{2} \times 1 \times 5 \times 5$

$\longrightarrow \sf s =\: 25 + \dfrac{1}{2} \times 5 \times 5$

$\longrightarrow \sf s =\: 25 + \dfrac{1}{2} \times 25$

$\longrightarrow \sf s =\: 25 + \dfrac{25}{2}$

$\longrightarrow \sf s =\: 25 + 12.5$

$\longrightarrow \sf s =\: 25 + \dfrac{125}{10}$

$\longrightarrow \sf s =\: \dfrac{250 + 125}{10}$

$\longrightarrow \sf s =\: \dfrac{375}{10}$

$\longrightarrow \sf\bold{\red{s =\: 37.5\: m}}$

$\therefore$ The distance covered by the car in that time is 37.5 m .