Question

a bus starting from rest acceleration in a straight line at a constant rate of 5 m/s² for 12 seconds. calculate the distance travelled by the bus during this time interval.​

Answer 1

Answer:

360 m

Explanation:

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

• Initial velocity (u) = 0 m/s (As it starts from rest)
• Acceleration (a) = 5 m/s²
• Time taken (t) = 12 seconds

We are asked to calculate the distance travelled by the bus during this time interval.

By using the second equation of motion :

⇒ s = ut + ½at²

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

⇒ s = (0 × 12) + ½ × 5 × (12)²

⇒ s = ½ × 5 × 144

⇒ s = 1 × 5 × 72

⇒ s = 360 m

∴ The distance travelled by the bus during this time interval is 360 m.

More Information :

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

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