Question

Starting from rest a racing
car accelerated
at 8m/6m. Calculate it velocity after covering
289 m ​

Answer 1

Appropriate Question:

Starting from rest a racing car accelerated at 8 m/s². Calculate it velocity after covering 289 m.

Answer:

68 m/s

Explanation:

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

• Initial velocity (u) = 0 m/s
• Acceleration (a) = 8 m/s²
• Distance covered (s) = 289 m

We are asked to calculate its velocity after covering 289 m i.e, its final velocity (v) .

By using the third equation of motion :

⇒ v² - u² = 2as

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

⇒ v² - (0)² = 2 × 8 × 289

⇒ v² - 0 = 4624

⇒ v² = 4624 + 0

⇒ v² = 4624

⇒ v = √4624

⇒ v = 68 m/s

∴ Its velocity after covering 289 m is 68 m/s.

$\rule{200}2$

More Information :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + at \quad}} \\ \\ \pmb{\sf{ \quad \: s= ut + \cfrac{1}{2}a{t}^{2} \quad \: } } \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2as \quad \:}}\end{array}}$

These are the three equations of motion.

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