An aeroplane accelerates uniformly from an initial velocity of 100 m.s-1 to a velocity of 200 m.s-1 in 10 seconds. It then accelerates uniformly to a final velocity of 240 m.s-1 in 20 seconds. Let the direction of motion of the aeroplane be the positive direction.

(a) Calculate the acceleration of the aeroplane during the first 10 seconds of the motion

(b) Calculate the distance the plane travels in the first 10 seconds

(c) Calculate the acceleration of the aeroplane during the next 20 seconds of its motion.

(d) Calculate the distance the plane travels in 20 seconds

[tex]\boxed{\begin{minipage}{9 cm}\underline{The Constant Acceleration Equations (SUVAT)}\\\\s = displacement in m (meters)\\u = initial velocity in m s$^{-1}$ (meters per second)\\v = final velocity in m s$^{-1}$ (meters per second)\\a = acceleration in m s$^{-2}$ (meters per second per second)\\t = time in s (seconds)\\\\When using SUVAT, assume the object is modeled\\ as a particle and that acceleration is constant.\end{minipage}}[/tex]

Part (a)

Given:

u = 100 m s⁻¹
v = 200 m s⁻¹
t = 10 s

[tex]\begin{aligned}\textsf{Using }\:\:v & = u+at\\\\\sf \implies 200 & = \sf 100+10a\\\sf 200-100 & = \sf 10a\\\sf 10a & = \sf 100\\ \sf a & = \sf 100 \div 10\\\sf a & = \sf 10 \:\: m\:s^{-2}\end{aligned}[/tex]

Part (b)

Given:

u = 100 m s⁻¹
v = 200 m s⁻¹
t = 10 s

[tex]\begin{aligned}\textsf{Using }\:\: s & = \dfrac{1}{2}(u+v)t\\\\\implies s & = \sf \dfrac{1}{2}(100+200)(10)\\& = \sf 5(300)\\& = \sf 1500 \:\: m\end{aligned}[/tex]

Part (c)

Given:

u = 200 m s⁻¹
v = 240 m s⁻¹
t = 20 s

[tex]\begin{aligned}\textsf{Using }\:\:v & = u+at\\\\\sf \implies 240& = \sf 200+20a\\\sf 240-200 & = \sf 20a\\\sf 20a & = \sf 40\\ \sf a & = \sf 40\div 20\\\sf a & = \sf 2\:\: m\:s^{-2}\end{aligned}[/tex]

Part (d)

It is not clear from the question if the distance the plane travels is:

Distance in the next 20 seconds, or
Distance in the first 20 seconds.

Therefore, I have calculated both.

Distance the plane travels in the next 20 seconds

(i.e. from 10 to 30 seconds)

Given:

u = 200 m s⁻¹
v = 240 m s⁻¹
t = 20 s

[tex]\begin{aligned}\textsf{Using }\:\: s & = \dfrac{1}{2}(u+v)t\\\\\implies s & = \sf \dfrac{1}{2}(200+240)(20)\\& = \sf 10(440)\\& = \sf 4400\:\: m\end{aligned}[/tex]

Distance the plane travels in the first 20 seconds

This is the sum of the distance the plane travels in the first 10 seconds (from part b) and the distance the plane travels in the next 10 seconds (so between 10 and 20 seconds).

Given:

u = 200 m s⁻¹
a = 2 m s⁻²
t = 10 s

[tex]\begin{aligned}\textsf{Using }\:\: s & = ut+\dfrac{1}{2}at^2\\\\\implies s & = \sf (200)(10)+ \dfrac{1}{2}(2)(10)^2\\& = \sf 2000+100\\& = \sf 2100\:\: m\end{aligned}[/tex]

Therefore, the distance the plane travels in the first 20 seconds of its journey is:

= 1500 m + 2100 m = 3600 m

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