Question

A car travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of -0.5 m/s^2. Find how far the car will go before it is brought to rest

Answer 1

Answer:

625 m

Explanation:

As per rhe provided information in the given question, we have :

• Initial velocity (u) = 90 km/h
• Acceleration (a) = –0.5 m/s²
• Final velocity (v) = 0 m/s (Since, it brought to rest)

We are asked to calculate the distance it shall cover before it is brought to rest.

Before commencing the steps, let's first convert the units of initial velocity in its standard form,that is m/s.

→ Initial velocity = 90 km/h

• In order to convert km/h to m/s, we multiply the values with 5/18.

→ Initial velocity = (90 × $\sf \dfrac{5}{18}$) m/s

→ Initial velocity = (5 × 5) m/s

→ Initial velocity = 25 m/s

Therefore, initial velocity in m/s is 25 m/s.

Now, we need to calculate the distance it shall cover before it is brought to rest. In order to calculate the distance, we'll be using the third equation of motion.

By using the third equation of motion,

★ v² - u² = 2as

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

Substituting the values,

→ (0)² - (25)² = 2 × (-0.5) × s

→ -625 = -1 × s

→ -625 ÷ (-1) = s

→ 625 metres = Distance

Therefore, the car will go 625 m before it is brought to rest.