Question

As an airplane is taking off at an airport its position is closely monitored by radar. The following three positions are measured with their corresponding times:
x₁ = 264.08 m at t₁ = 4.90 s,
x₂ = 312.33 m at t₂ = 5.40 s,
x₃ = 364.72 m at t₃ = 5.90 s.
What is the acceleration of the airplane at t₂ = 5.40 s? (Assume that the acceleration of the airplane is constant.)​

a=_______m/s^2

Answer 1

Let the initial position of the airplane be x1 = 264.08 m

Time when position x1 is noted is t1 = 4.90 s

Next position noted is x2 = 312.33 m

Corresponding time is t2 = 5.40 s

Last reading about position of airplane is x3 = 364.72 m

Corresponding time is t3 = 5.90 s

We can observe that time interval between the readings is ∆t = 0.5 s

Actually we can take that airplane is at rest initially and time is equal to zero. We can use second kinematical equation

S = ut + 1/2 a t² in the form

x1 = u t1 + 1/2 at1²

to get the acceleration by setting initial speed u = 0

But it is not the case here. We have to calculate the acceleration between the times t1 and t3 using the information provided.

We solve the problem by finding the instantaneous velocities in the two cases and find acceleration based on it.

Let the instantaneous velocity between the times t1 and t2 be v1

Velocity v = displacement / time   

So v1 = ( x2 - x1 ) / ( t2 - t1 )

= ( 312.33 - 264.08 ) /( 5.4 - 4.9 )

= 48.25 / 0.5

= 96.50 m/s

This can be taken as velocity of airplane at time t2 = 5.40 s

Similarly instantaneous velocity between the times t2 and t3 be v2

v2 = ( x3 - x1 ) / ( t3 - t2 )

= ( 364.72 - 312.33 ) / ( 5.90 - 5.40 )

= 52.39 / 0.5

= 104.78 m/s

Now acceleration is given by

a = change in the velocity / time interval

= dv / dt

= ( v2 - v1 ) / ( t3 - t2 )

= ( 104.78 - 96.50 ) / ( 5.9 - 5.4 )

= 8.28 / 0.5

= 16.56 m/s²

As acceleration is uniform , it will have same value at time t = 5.40 s

So acceleration of the airplane is a = 16.56 m/s²