Question

The velocity vs s graph for an airplane travelling on a straight run way is shown determine the acceleration of the plane at s= 50 m and s= 150 m

Answer 1

Given:

Velocity vs Displacement graph for an aeroplane travelling on a straight runway has been shown.

To find:

Acceleration of the plane at:

• s = 50 m
• s = 150 m

Calculation:

Acceleration is defined as the instantaneous rate of change of velocity with respect to time. Mathematically, it is written as:

$\therefore \: a = \dfrac{dv}{dt}$

$= > \: a = \dfrac{dv}{dx} \times \dfrac{dx}{dt}$

$= > \: a = v \times \dfrac{dv}{dx}$

$\boxed{ \sf{ = > \: a = v \times (slope \: of \: v - s \: \: graph)}}$

At s = 50 m, the velocity will be :

$\therefore \: \dfrac{50}{100} = \dfrac{v}{40}$

$= > \: v = 20 \: m {s}^{ - 1}$

So, acceleration will be :

$= > \: a = v \times \dfrac{dv}{dx}$

$= > \: a = 20 \times \dfrac{(20 - 0)}{(50 - 0)}$

$= > \: a = 20 \times \dfrac{20}{50}$

$\boxed{ = > \: a = 8 \: m {s}^{ - 2} }$

At s = 150 m, the velocity will be:

$\therefore \: \dfrac{150}{200} = \dfrac{v}{50}$

$= > \: v = 45 \: m {s}^{ - 1}$

So, acceleration will be:

$= > \: a = v \times \dfrac{dv}{dx}$

$= > \: a = 45 \times \dfrac{(50 - 40)}{(200 - 100)}$

$= > \: a = 45 \times \dfrac{10}{100}$

$\boxed{ = > \: a = 4.5 \: m {s}^{ - 1} }$

Hope It Helps.

Answer 2

8ms-²

REMEMBER!!V= ds/dt

Same way with 150 m(THATS UR H.W)

HAVE A GOOD DAY/NIGHT