Question

If an airplane has a takeoff speed of 95 m/s and requires 1390 m to reach that speed, what is the acceleration of the plane and the time required to reach this speed?

Answer 1

To find the acceleration of the airplane, we can use the formula:

$v=u+at$

Where:

• $v$ is the final velocity (takeoff speed) = 95 m/s
• $u$ is the initial velocity (starting from rest) = 0 m/s
• $a$ is the acceleration of the airplane (what we need to find)
• $t$ is the time required to reach the final velocity (what we also need to find)

Since the airplane starts from rest, the initial velocity ($u$) is 0 m/s. Therefore, the formula simplifies to:

$v=at$

Plugging in the values we have:

$95=a\times t$

To find the time required ($t$) to reach the takeoff speed, we need another equation. The formula to calculate distance traveled ($s$) under constant acceleration is:

$s=ut+\frac{1}{2}a{t}^2$

Where:

• $s$ is the distance traveled = 1390 m
• $u$ is the initial velocity = 0 m/s
• $a$ is the acceleration of the airplane (what we need to find)
• $t$ is the time required to reach the final velocity (what we also need to find)

Plugging in the values we have:

1390=0×t+12at21390 = 0 imes t + frac{1}{2}at^21390=0<span