Question

2. An airplane travels 500 m down the runway before taking off. If it starts from rest, moves with constant acceleration, and becomes airborne in 20 s, with what velocity does it take off?

3. A car moving with constant acceleration covers the distance between two points 60 m apart in 6 s. Its velocity as it passes the second point is 15 m/s.
a) What is its velocity at the first point?
b) What is the acceleration?

4. A triumph TR-7 starts at rest and accelerates to a speed of 27 m/s in 11.8 s. (a) Calculate the car’s acceleration. (b) Calculate the distance traveled during this time.

Answer 1

2. The airplane started from rest has 500 m and 20 s to become airborne. So, the final velocity is

$$v_f=\frac{2d}{t}=\frac{2\cdot500}{20}=50\text{ m/s}.$$

3. We know that in uniformly accelerated motion, distance increases with time according to the following law:

$$d=v_0t+\frac{at^2}{2},\rightarrow v_0=\frac{d}{t}-\frac{at}{2}.$$

Also, we know that

$$a=\frac{v_f^2-v_0^2}{2d}=\frac{v_f^2-(\frac{d}{t}-\frac{at}{2})^2}{2d},\\\space\\ a=1.67\text{ m/s}.\\ v_0=\sqrt{v_f^2-2ad}=5\text{ m/s}.$$

4. Acceleration:

$$a=\frac vt=\frac{27}{11.8}=2.29\text{ m/s}^2,\\\space\\ d=\frac{at^2}{2}=159\text{ m}.$$