Back to Course

SS2: PHYSICS - 1ST TERM

0% Complete
0/0 Steps
  1. Scalars & Vectors | Week 1
    5 Topics
    |
    1 Quiz
  2. Equations of Motion | Week 2
    3 Topics
    |
    1 Quiz
  3. Projectile | Week 3
    5 Topics
  4. Equilibrium of Forces I | Week 4
    4 Topics
  5. Equilibrium of Forces II | Week 5
    4 Topics
  6. Stability of a Body | Week 6
    4 Topics
    |
    1 Quiz
  7. Simple Harmonic Motion (SHM) | Week 7
    4 Topics
  8. Speed, Velocity & Acceleration & Energy of Simple Harmonic Motion | Week 8
    5 Topics
    |
    1 Quiz
  9. Linear Momentum | Week 9
    6 Topics
    |
    1 Quiz
  10. Mechanical Energy & Machines | Week 10
    2 Topics
    |
    1 Quiz



  • Follow us

Lesson 2, Topic 3
In Progress

Application of Equations of Motion

Lesson Progress
0% Complete

Let’s list the four equations of motions again;

(1) v = u + at

(2) s = \(\scriptsize ut \: +\: \normalsize \frac{at^2}{2} \)

(3) \( \scriptsize v^2 = u ^2 \: + \: 2as \)

(4) s = \( \frac{u\: + \:v}{2} \scriptsize\: \times\: t \)

Also, the equations of motion for the uniformly accelerated bodies;

(1) v = \( \scriptsize u \pm gt \)

(2) h = \( \scriptsize ut \pm \normalsize \frac{1}{2} \scriptsize gt^2 \)

(3) v2 = \( \scriptsize u^2 \pm 2gh \)

Example 1:

A bike travels from rest and accelerates uniformly to a velocity of 20ms-1 in 5 seconds. Calculate the distance travelled.

Solution:

u = 0. v = 20ms-1, t = 5s

s = \( \frac{u \: + \: v}{2} \scriptsize \times t \)

s = \( \frac{0 \: + \: 20}{2} \scriptsize \times 5 \)

   = 10 x 5

S = 50m

Example 2:

A ball thrown vertically upward with an initial velocity of 40 m/s has a deceleration of 10m/s2 Calculate its
(a) velocity after 2.5s and
(b) time to reach maximum height. (g = 10m/s2.)

Solution:

Values given

Initial velocity u = 40m/s; time t = 2.5s; velocity at time 2.5s = ?; deceleration = 10m/s2

(a) You have to choose a formula that contains all the values given in the question.

Using v = u + gt

Substitute in the values

v = 40 + (-10 x 2.5)

v = 40 – 25

v = 15m/s

(b) Time taken to reach maximum height.

To calculate the time taken to reach maximum height, you must remember that at maximum height, final velocity v = 0m/s.

Using v = u + gt

Substitute in the values

0 = 40 + (-10t)

-40 = -10t

t = \( \frac{-40}{-10} \)

t = 4s

Example 3:

A van accelerates uniformly at 2m/s2 for 15s from an initial velocity of 5m/s. Calculate the distance travelled. (g = 10m/s2)

Solution:

Values given

Acceleration = 2m/s2; u = 5m/s; t = 15s; g = 10m/s2; s = ?

From the values we use the formula

s = \(\scriptsize ut \: +\: \normalsize \frac{at^2}{2} \)

Substitute in the values given;

s = \(\scriptsize 5 \: \times \: 15 \: +\: \normalsize \frac{2\: \times \: 15^2}{2} \)

s = \(\scriptsize 75 \: + \: \normalsize \frac{450}{2} \)

s = \(\scriptsize 75 \: + \: 225\)

s = 300m

Example 4:

A cup is thrown vertically upward from the ground with a velocity of 15m/s. (g = 10m/s2.). calculate:
(a) The maximum height reached
(b) The time to reach the maximum height
(c) The time to reach the ground again
(d) The velocity reached halfway up

Solution:

Recall that at maximum height, v = 0m/s

Values given

u = 15 m/s; v = 0 m/s; g = 10 m/s2.

(a) To calculate the maximum height reached, use the formula

v2 = u2 + 2as

Substitute given values into the formula

02 = 152 + 2 (-10) s

02 = 225 + -20s

-225 = -20s

s = \( \frac{-225}{-20} \)

s = 11.25m

(b) To calculate time to reach the maximum height we are going to use the value we calculated for the maximum height, s, previously.

Based on the values we have we are going to use the formula

s = \(\scriptsize ut \: +\: \normalsize \frac{at^2}{2} \)

Substitute in the values

11.25 = \(\scriptsize 15 \: \times \: t \: +\: \normalsize \frac{(-10)t^2}{2} \)

11.25 = \(\scriptsize 15 \: \times \: t \: +\: (-5)t^2 \)

11.25 = \(\scriptsize 15t \: -\: 5t^2 \)

0 = \(\scriptsize 15t \: -\: 5t^2 \: – \: 11.25 \)

-5t^2 + 15t – 11.25 = 0

Using the quadratic formula

⇒ \( \scriptsize t \; = \normalsize \frac{-b\pm\sqrt{b^2\; -\; 4ac}}{2a} \)

Where,

A = – 5; b = 15; c = -11.25

⇒ \( \scriptsize t \: = \normalsize \frac{-15\pm\sqrt{15^2\: -\: 4(-5)(-11.25)}}{2(-5)} \)

⇒ \( \scriptsize t \: = \normalsize \frac{-15\pm\sqrt{225\: -\: 225}}{-10} \)

⇒ \( \scriptsize t \: = \normalsize \frac{-15\pm\sqrt{0}}{-10} \)

⇒ \( \scriptsize t \: = \normalsize \frac{-15}{-10} \)

⇒ \( \scriptsize t \: = 1.5s \)

Time to reach maximum height = 1.5s

(c) The time to reach the ground again:

The time to reach the ground is the same as the time taken to reach maximum height.

Therefore,

Time to reach the ground = 2 x time taken to reach maximum height.

Time to reach the ground = 2 x 1.5s

Time to reach the ground = 3s.

(d) The velocity reached halfway up:

To calculate the velocity of the object halfway up, the time taken to travel half the distance is half of the time taken to reach the maximum height.

Or you use half the distance travelled to calculate its velocity half way up.

Therefore,

The distance half way up = \( \frac{11.25}{2} \) = 5.625m

Use the formula

v2 = u2 + 2as

v2 = 152 + 2(-10)5.625

v2 = 225 + (-112.5)

v2 = 225 – 112.5

v2 = 112.5

Take the square root of both sides

v = \( \scriptsize \sqrt{112.5} \)

v = 10.6 m/s

Responses

Your email address will not be published. Required fields are marked *

error: