Question

A jet accelerates down a runway at 4.20 m/s2 for 34.2 s until it finally lifts off the ground. What is the distance traveled before takeoff.

Answer 1

Answer:-

$\blue{\bigstar}$$\large\boxed{\rm\orange{2456.244 \: m}}$

• Given:-

Initial velocity [u] = 0

Time [t] = 34.2 sec.

Acceleration [a] = 4.20 ms-²

• To Find:-

Distance travelled [s] = ?

• Solution:-

• Using 2nd Equation of motion:-

$\pink{\bigstar}$$\large\boxed{\bf\purple{s = ut + \dfrac{1}{2} at^{2}}}$

here,

$\red{\dag}$ u = 0

$\red{\dag}$ t = 34.2

$\red{\dag}$ a = 4.20 ms-²

$\dashrightarrow$ $\bf{s}\sf{ = 0 \times 34.2 + \dfrac{1}{2} \times 4.20 \times (34.2)^{2}}$

$\dashrightarrow$$\bf{s}\sf{ = 0 + \dfrac{1}{2} \times 4.20 \times 1169.64}$

$\dashrightarrow$$\bf{s}\sf{ = 0 + \dfrac{1}{2} \times 4912.488}$

$\dashrightarrow$$\bf\green{s = 2456.244 \: m}$

Therefore, the distance travelled before takeoff is 2456.244 m.

Answer 2

Answer:

We can solve the given problem by using the following SUVAT equation :]

• S = ½ at²

Where, S is Distance travelled, a is the Acceleration of the Jet and t is the Time.

We've have given that, Acceleration (a) = 4.20 m/s² and Time (t) = 34.2 seconds.

• So,we can find the Distance travelled by the Jet by Substituting the given Data into the above given formula :]

$\sf => S = \dfrac{1}{2} \times 4.20 \times 34.2$

$\sf => S = 71.82 \: meter$