Question

An airplane starts its takeoff by moving at 1.7 m/s. The plane must then accelerate at 3.0 m/s2 for the next 30s to reach its final takeoff speed. To the nearest tenth of a kilometer, how far does the plane travel while accelerating to its takeoff speed?

Answer 1

Answer:

The distance is 1.4 km.

Explanation:

Given that,

Initial velocity u= 1.7 m/s

Accelerate a= 3.0 m/s²

Time = 30 s

We need to calculate the distance

Using equation of motion

$s= ut+\dfrac{1}{2}at^2$

Where, u = initial velocity

a = acceleration

t = time

Put the value in the equation

$s=1.7\times30+\dfrac{1}{2}\times3.0\times(30)^2$

$s=1401\ m$

$s=1.4\ km$

Hence, The distance is 1.4 km.