Question

a plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. determine the acceleration of the plane and the time required to reach this speed

Answer 1

Answer on Question#78929 - Physics - Other

A plane has a takeoff speed of $88.3 \, \text{m/s}$ and requires $1365 \, \text{m}$ to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

Solution:

The final and initial velocities $v_{f}, v_{i}$, acceleration $a$ and distance $d$ are related by the following expression

$$d = \frac {v _ {f} ^ {2} - v _ {i} ^ {2}}{2 a}$$

In our case $v_{f} = 88 \, \text{m/s}$, $v_{i} = 0 \, \text{m/s}$, $d = 1365 \, \text{m}$, thus we have

$$a = \frac {v _ {f} ^ {2} - v _ {i} ^ {2}}{2 d} = \frac {\left(8 8 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2} - \left(0 \frac {\mathrm {m}}{\mathrm {s}}\right) ^ {2}}{2 \cdot 1 3 6 5 \mathrm {m}} = 2. 8 4 \frac {\mathrm {m}}{\mathrm {s} ^ {2}}$$

The final and initial velocities $v_{f}, v_{i}$, acceleration $a$ and time $t$ are related by the following expression

$$v _ {f} - v _ {i} = a t$$

Thus the time required to reach this speed:

$$t = \frac {v _ {f} - v _ {i}}{a} = \frac {8 8 \frac {\mathrm {m}}{\mathrm {s}}}{2 . 8 4 \frac {\mathrm {m}}{\mathrm {s} ^ {2}}} = 3 1 \mathrm {s}$$

Answer: acceleration: $2.84\frac{\mathrm{m}}{\mathrm{s}^2}$, time: 31 s.

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