Question

A car increases its velocity from 72km/h to 108km/h in 10 seconds. its acceleration is​

Answer 1

Answer:

1 m/s²

Explanation:

Given:

• Initial velocity = u = 72 km/h
• Final velocity = v = 108 km/hr
• Time taken = t = 10 seconds

To find:

• Acceleration of the car

Using the first equation of motion:

V=u+at

Converting the initial velocity into m/s = 72×5/18 = 20 m/s

Converting the final velocity into m/s = 108×5/18= 30 m/s

Substituting the values:

30=20+a×10

30-20=10a

10=10a

a = 10/10

a = 1 m/s²

The acceleration of the car is equal to 1 m/s²

Answer 2

✰ Given -

• Initial Velocity (u) = 72 km/h
• Final Velocity (v) = 108 km/h
• Time (t) = 10 sec

✰ To Find -

• Acceleration = ?

✰ Solution -

Let's covert some measurements,

• $⟶72 \times \frac{5}{18}$

$⟶20m/s$

• $⟶108 \times \frac{5}{18}$

$⟶30m/s$

We know 1st Equation of Kinematics,

$✓ \: v = u + at$

Here,

• V = Final Velocity (20m/s)
• U = Initial Velocity (30m/s)
• A = Acceleration
• T = Time (10sec)

Placing Values,

$⟶20 = 30 + a \times 10$

$⟶30 - 20 = 10a$

$⟶10 = 10a$

$⟶a = \frac{\cancel{10}}{\cancel{10}}$

$⟶a = 1 {m/s}^{2}$

⛬ Acceleration is 1m/s^2