Question

A jet aircraft launched from an aircraft carrier’s accelerated from rest along a 94-m track for 2.5 s (a) what is the acceleration of the aircraft, assuming it is constant? (B) what is the launch speed of the jet?

Answer 1

Answer:

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Answer 2

Given:

Distance travelled by jet aircraft $=94 m$

Time taken $=2.5 s$

To Find:

(a) acceleration

(b) launch speed

Solution:

(a)

From equations of motions, we have,

$\[s = ut + \frac{1}{2}a{t^2}\]$                               ... (i)

where, $s=$ distance; $u=$ initial velocity; $a=$ acceleration; $t=$ time

Therefore, the acceleration of the aircraft can be calculated as

$\[94m = 0 \times 2.5s + \frac{1}{2}a{(2.5s)^2}\]$        ($u=0$ as aircraft is accelerated from rest)

$\[a = \frac{{94m \times 2}}{{{{(2.5s)}^2}}}\]$

$\[a = 30.08m/{s^2}\]$

Hence, the acceleration of aircraft is $\[30.08m/{s^2}\]$.

(b)

Launch velocity of the jet aircraft can be given as

$\[v = at\]$

$\[v = 30.08m/{s^2} \times 2.5s\]$

$\[v = 75.2m/s\]$

Hence, the launch speed of the jet is $\[75.2m/s\]$.