Question

A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar

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Answer 1

Appropriate Question :

• A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.

Given :

• Initial Velocity (u) = 108 km/hr
• Time taken (t) = 4 seconds
• Final Velocity (v) = 0 m/s
• mass of motorcar (m) = 1000 kg

To Find :

• The force (F) exerted by the brakes on the motorcar = ?

Solution :

★ First of all convert initial velocity (u) from km/hr to m/s :

→ Initial Velocity = 108 km/hr

→ Initial Velocity = 108 × (5 ÷ 18)

→ Initial Velocity = 6 × 5

→ Initial Velocity = 30 m/s

• Hence,the initial velocity (u) of motorcar is 30 m/s.

★ Now, let's find the acceleration of the motorcar :

➝ Acceleration = (Final Velocity - Initial Velocity) ÷ time taken

➝ Acceleration = (0 - 30) ÷ 4

➝ Acceleration = -30 ÷ 4

➝ Acceleration = -7.5 m/s²

• Hence, acceleration (a) of the motorcar is -7.5 m/s².

★ Finding force exerted (F) by the brakes on the motorcar :

⇒ Force = mass × acceleration

⇒ Force = 1000 × -7.5

⇒ Force = -7500 Newtons

• Hence,the force exerted (F) by the brakes on the motorcar is -7500 Newton's.

Answer 2

$\huge\underline\mathcal\orange{QUESTION}$

A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar.

$\huge\underline\mathcal\red{ANSWER}$

$u = 108 \: km \: h \\ converting \: the \: units \\ u = 108 \times \frac{5}{18} = 30 \: m \: s$

$v = 0 \: m \: s \\ t = 4 \: sec \\ formula \:for \: acceleration \: \\ acc = \frac{v - u}{t} \\ inserting \: the \: values \\ acc = \frac{0 - 30}{4} = - 3.75 {m \: s}^{2} (retardation)$

$formula \: for \: force =f = ma \\ f = - 3.75 \times 1000=-7500 N$

$\huge\underline\mathcal\purple{BTS \: EXO}$