Question

An airplane speeds up from rest at constant rate of 12m/s2. How long dies it take to cover a distance of 1.50 x 102m

Answer 1

Given :

• Acceleration of the Airplane (a) = 12 m/s²

• Distance covered by it (S) = 1.50 × 10² m.

To find :

The time taken to cover the distance of 1.50 × 10² m. (t)

Solution :

Since , we are provided with the distance covered and the acceleration produced , we can use the second Equation of Motion to find the time taken.

But according to the given information , it is being stated that the airplane is being Started from the rest , i.e, the initial velocity of the airplane will be 0 , although the final Velocity will have some value.

Here , we concluded that :

Initial velocity of the airplane is 0 m/s.

We know the second Equation of Motion :

$\boxed{\bf{S = ut + \dfrac{1}{2}at^{2}}}$

Where :

• S = Distance traveled
• u = Initial Velocity
• t = Time Taken
• a = Acceleration

From the second Equation of Motion , we get : (When , u = 0)

$:\implies \bf{S = ut + \dfrac{1}{2}at^{2}}$

$:\implies \bf{S = 0 \times t + \dfrac{1}{2}at^{2}}$

$:\implies \bf{S = \dfrac{1}{2}at^{2}}$

$\boxed{\therefore \bf{S = \dfrac{1}{2}at^{2}}}$

Hence, the Second equation of motion , when Initial Velocity is 0 (i.e, u = 0) is $\bf{\dfrac{1}{2}at^{2}}$

Now using the second Equation of Motion when Initial velocity is 0, and substituting the values in it, we get :

$:\implies \bf{S = \dfrac{1}{2}at^{2}}$

$:\implies \bf{1.5 \times 10^{2} = \dfrac{1}{2} \times 12 \times t^{2}}$

$:\implies \bf{1.5 \times 10^{2} = 6 \times t^{2}}$

$:\implies \bf{\dfrac{1.5 \times 10^{2}}{6} = t^{2}}$

$:\implies \bf{\dfrac{150}{6} = t^{2}}$

$:\implies \bf{25 = t^{2}}$

$:\implies \bf{\sqrt{25} = t}$

$:\implies \bf{5 = t}$

$\boxed{\therefore \bf{t = 5\:s}}$

Hence the time taken to cover a distance of 1.5 × 10² m with a acceleration of 12 m/s² is 5 s.