Question

A car moving at 36 km/h is brought to rest in 0.1 km. What is the retardation?​

Answer 1

Answer:

36km/h=10m/s

Explanation:

u=36

v=0

s=0.1 so we use third law of motion

${v}^{2} = {u}^{2} + 2as \\ {0}^{2} = {10}^{2} + 2 \times a \times 0.1 \\ 0 = 100 + 0.2a \\ 0.2a = - 100 \\ a = - 100 \div 0.2 \\ a = - 500m \div s$

Answer 2

Answer:

$\large{\underline{\boxed{\sf Retardation = 0.5 \: m/s^2}}}$

Explanation:

Given that ;

Initial velocity (u) = 36 km/h

Converting km/h into m/s.

⇒ u = 36 * 5/18

⇒ u = 10 m/s

Final velocity (v) = 0 m/s.

Distance travelled (s) = 0.1km

⇒ s = 0.1 km = 100 m

We need to find retardation (-a).

To find it, we will use third equation of motion ;

v² - u² = 2as

⇒ 0 - 10² = 2 * a * 100

⇒ - 100 = 200a

⇒ a = $\sf - \dfrac{100}{200}$

⇒ a = - 0.5 m/s²

⇒ - a = 0.5 m/s²

Hence, the answer is 0.5 m/s².