18. Motion with variable acceleration I

From Mechanics

18. Motion With Variable Acceleration I

Key Points

Example 18.1

A ball is thrown vertically upwards. The height, h metres, of the ball at time t seconds, is given by:

h=1.5+7t−4.9t2

Show that the acceleration of the ball is constant and find its maximum height.

Solution

Differentiate once to find the velocity:

v=dtdh=7−9.8t

Differentiate again to find the acceleration:

a=dtdv=−9.8

Hence the acceleration is constant.

At its maximum height the velocity of the ball will be zero.

7−9.8t=0t=79.8=0.714 s (to 3sf)

This time can be substituted to find the maximum height:

h=1.5+7798−4.979.82=4 m  

Example 18.2

A particle moves so that its position vector, r metres, at time t seconds is given by:

r=t2−6t+9i+2t+6j 

The unit vectors i and j are directed east and north respectively.

a) Find the velocity and acceleration the ball.

b) Find the time when the particle is moving due north.

Solution

a) Differentiate the position vector once to find the velocity:

v=dtdr=2t−6i+2j 

Differentiate the velocity to find the acceleration.

a=dtdv=2i

b) When travelling north the i component of the velocity will be zero. This leads to:

2t−6=0t=3

The particle will be moving north after 3 seconds.

Example 18.3

A particle, of mass 8 kg, moves so that at time t its position vector is r, where

r=(8t−t2)i+(4+12t−t2)j

and the unit vectors i and j are perpendicular.

a) Find the velocity and acceleration of the particle.

b) Find the magnitude of the force acting on the particle.

c) Find the time when the speed of the particle is 4 ms−1.

Solution

a) Differentiate the position vector once to obtain the velocity of the particle.

v=dtdr=(8−2t)i+(12−2t)j

Differentiate the velocity to obtain the acceleration:

a=dtdv=−2i−2j

b) The resultant force on the particle is found using F=ma. Hence:

F=8(−2i−2j)=−16i−16j N

Now the magnitude, F, of this force can be found.

F=(−16)2+(−16)2=226 N (to 3sf) 

c) First find an expression for the speed, v, of the particle. This is given by:

v=(8−2t)2+12−2t2 

Given that the speed is 4 \displaystyle \text{m}{{\text{s}}^{-1}} leads to an equation which can be solved as shown below:

\displaystyle \begin{align} & 4=\sqrt{{{(8-2t)}^{2}}+{{\left( 12-2t \right)}^{2}}} \\ & {{4}^{2}}={{(8-2t)}^{2}}+{{\left( 12-2t \right)}^{2}} \\ & 16=64-32t+4{{t}^{2}}+144-48t+4{{t}^{2}} \\ & 0=8{{t}^{2}}-80t+192 \\ & 0={{t}^{2}}-10t+24 \\ & 0=(t-6)(t-4) \\ & t=6\text{ or }t=4 \\ \end{align}

The speed will be 4 \displaystyle \text{m}{{\text{s}}^{-1}} when t = 4 and when t = 6.