Question

An airplane, taking off from a field, has a run of 500 m. What is the acceleration if he leaves the ground in 10 s from the start? Also find the take-off velocity.

Answer 1

Step 1: Given data

1. Distance traveled by airplane, s = 500 m.
2. Required time, t = 10 sec.
3. Initial velocity, u = 0 m/s.

Let, the acceleration and velocity of the airplane be a and v respectively.

Step 2: Finding the acceleration of the plane

From the first second equation of motion we can write;

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

where, s = distance , u = initial velocity, t = time, v = final velocity, a = acceleration.

From above equation.

$$\begin{gathered} s=\left(0\times t\right)+\frac{1}{2}a{t}^2 \\ s=\frac{1}{2}a{t}^2 \\ a=\frac{2s}{t^2} \\ a=\frac{2\times 500}{{\left(10\right)}^2}=\frac{1000}{100} \\ a=10m/s^2. \end{gathered}$$

Therefore the acceleration of the airplane after 10 sec will be $10m/s^{2.}$

Step 2: Finding the velocity of the plane

From the first equation of motion, we can write

$v=u+at$

From the above equation,

$$\begin{gathered} v=0+at \\ v=10\times 10=100 \\ v=100m/s. \end{gathered}$$

Therefore, the take-off velocity of the airplane is 100 m/s.

Hence, the acceleration of the airplane after 10 sec and the take-off velocity of the airplane will be $10m/s^{2.}$ and 100 m/s respectively.