Question

A car initially at rest starts moving with a constant acceleration of 0.5ms^-2 and travels a distance 25 cm .calculate its final velocity and time taken.​

Answer 1

Answer:

• Final velocity = 0.5m/s
• Time taken = 1 second

Explanation:

According to the Question

It is given that

• Initial velocity ,v = 0m/s
• Acceleration ,a = 0.5m/s²
• distance covered ,s = 25cm

we need to calculate the final velocity and the time taken by the car .

Firstly we calculate the final velocity of the car .

Coversion of Unit

→ distance = 25/100m = 0.25m

Using Kinematics Equation

• v² = u² + 2as

Putting Value we get

➻ v² = 0² + 2×0.5 × 0.25

➻ v² = 0 + 1 × 0.25

➻ v ² = 0.25

➻ v = √0.25

➻ v = 0.5m/s

• Hence, the final velocity of the car is 0.5m/s.

Now, calculating the time taken .

Using Kinematics Equation

• v = u + at

substitute the value we get

➻ 0.5 = 0 + 0.5 × t

➻ 0.5 = 0.5 × t

➻ 0.5/0.5 = t

➻ t = 1 s

• Hence, the time taken by the car is 1 second.

Answer 2

⚘ Question :-

• A car initially at rest starts moving with a constant acceleration of 0.5 m/s² and travels a distance 25 cm. Calculate it's final velocity and time taken.

⚘ Answer :-

• Final velocity = 0.5 m/s
• Time taken = 1 second

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⚘ Given :-

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 0.5 m/s²
• Distance travelled = 25 cm

⚘ To Find :-

• Final velocity (v) = ?
• Time taken (t) = ?

⚘ Solution :-

★ Calculating final velocity ::

• Using third eqⁿ of motion.

We know that,

✪ v² = u² + 2as ✪

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance travelled

We have,

• u = 0 m/s
• a = 0.5 m/s²
• s = 25 cm
• s (In m) = (25/100) m = 0.25 m
• v = ?

By plugging all values in eqⁿ we get,

⇒ v² = (0)² + 2(0.5)(0.25)

⇒ v² = 0 + 2 × 0.5 × 0.25

⇒ v² = 1 × 0.25

⇒ v² = 0.25

⇒ v = √0.25

⇒ v = √(0.5 × 0.5)

➠ v = 0.5 m/s

∴ Final velocity is 0.5 m/s.

Now,

★ Calculating time taken ::

• Using second eqⁿ of motion.

We know that,

✪ v = u + at ✪

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time taken

We have,

• v = 0.5 m/s
• u = 0 m/s
• a = 0.5 m/s²
• t = ?

By plugging all values in eqⁿ we get,

⇒ 0.5 = 0 + (0.5)(t)

⇒ 0.5 = 0.5 × t

⇒ 0.5t = 0.5

⇒ t = 0.5/0.5

By canceling 0.5 with 0.5 we get,

⇒ t = 1/1$\:$

➠ t = 1 second

∴ Time taken is 1 second.

⚘ Learn More :-

✧ Three equations of motion ✧

1. v = u + at ㅤㅤㅤㅤㅤ[Used above]
2. s = ut + ½ at²
3. v² = u² + 2asㅤ ㅤ ㅤ[Used above]

✧ Some important definitions ✧

• Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

• Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

• Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.

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