Question

A plane has take off 300km/h what is acceleration in m/s^2 of plane started from rest and took 45seconds to take off ?​

Answer 1

Solution :

⏭ Given:

✏ Initial velocity of plane = 0

✏ Final velocity of plane = 300kmph

✏ Time interval = 45s

⏭ To Find:

✏ Acceleration of plane

⏭ Concept:

✏ Acceleration is defined as ratio of change in velocity to the time interval.

⏭ Calculation:

✏ Formula of acceleration in terms of change in velocity and time interval is given by

$\star \: \underline{ \boxed{ \bold{ \sf{ \pink{ \large{a = \dfrac{v - u}{t}}}}}}} \: \star$

⏭ Terms indication:

✏ a denotes acceleration

✏ v denotes final velocity

✏ u denotes initial velocity

✏ t denotes time interval

⏭ Conversation:

✏ 300kmph = 83.33mps

⏭ Calculation:

$\mapsto \sf \: a = \dfrac{83.33 - 0}{45} \\ \\ \mapsto \sf \: a = \dfrac{83.33}{45} \\ \\ \mapsto \: \underline{ \boxed{ \bold{ \sf{ \purple{ \large{a = 1.85 \: m {s}^{ - 2} }}}}}} \: \gray{ \bigstar}$

⏭ Additional information:

• Acceleration is vector quantity.
• Acceleration can be positive, negative and zero.