CHAPTER 2.1 LINEAR MOTION

Solution :

When s = 0

= ut − 1 gt 2 = 0
2
Learning Outcome:
t(30 − 5t) = 0
(a) derive and use
equations of then t = 0 and t = 6
motion with
constant given s = 25 m , u = 30 ms-1
acceleration;
1
(b) sketch and use s = ut − 2 gt 2
the graphs of
displacement Figure 2-
time, velocity-time
25 = 30t − 5t 2
and
acceleration-time (t − 1)(t − 5) = 0

for the motion of a The time interval will be 5s - 1s = 4s
body with constant

acceleration;

Example :

Learning Outcome: An object thrown upward from a point P, initial velocity 20 ms−1
that is 25 m from the surface of the earth. Sketch the graph of .
(a) derive and use (a) Displacement - time
equations of (b) Velocity - time
motion with (c) Speed – time
constant
acceleration; • Maximum Height, H : v = 0, s = H

(b) sketch and use v 2 = u2 − 2gs
the graphs of
displacement 0 = u2 − 2gH
time, velocity-time
and H = u2 = 400 = 20m
acceleration-time 2g 2(9.8)
for the motion of a
body with constant
acceleration;

Example :

Learning Outcome: An object thrown upward from a point P, initial velocity 20 ms−1
that is 25 m from the surface of the earth. Sketch the graph of .
(a) derive and use (a) Displacement - time
equations of (b) Velocity - time
motion with (c) Speed – time
constant
acceleration; • Assume s = 0 at Point P. Time to return to point P = tP , u = 20 ms-1

(b) sketch and use u sin g s = ut - 1 gt 2
the graphs of 2
displacement
time, velocity-time 0 = (20)t P − 1 gt 2
and 2 P
acceleration-time
for the motion of a t P (20 − 1 gtP ) = 0
body with constant 2
acceleration;
tP = 0 and tP = 4s

Example :

Learning Outcome: An object thrown upward from a point P, initial velocity 20 ms−1
that is 25 m from the surface of the earth. Sketch the graph of .
(a) derive and use (a) Displacement - time
equations of (b) Velocity - time
motion with (c) Speed – time
constant
acceleration; • from point P to ℎ surface ,

(b) sketch and use s = −25 m. Time taken to the earth surface , t = ts
the graphs of 1
displacement s = ut− 2 2
time, velocity-time
and −25 = (20) − 1 2
acceleration-time 2
for the motion of a
body with constant ( + 1)( − 5) = 0
acceleration;
∴ = 5

Example :

Learning Outcome: An object thrown upward from a point P, initial velocity 20 ms−1
that is 25 m from the surface of the earth. Sketch the graph of .
(a) derive and use (a) Displacement - time
equations of (b) Velocity - time
motion with (c) Speed – time
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Example :

Learning Outcome: An object thrown upward from a point P, initial velocity 20 ms−1
that is 25 m from the surface of the earth. Sketch the graph of .
(a) derive and use (a) Displacement - time
equations of (b) Velocity - time
motion with (c) Speed – time
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Example :

Learning Outcome: An object thrown upward from a point P, initial velocity 20 ms−1
that is 25 m from the surface of the earth. Sketch the graph of .
(a) derive and use (a) Displacement - time
equations of (b) Velocity - time
motion with (c) Speed – time
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;

Learning Outcome: THANK YOU

(a) derive and use Pear deck link
equations of https://app.peardeck.com/student/tmolknnxo
motion with
constant
acceleration;

(b) sketch and use
the graphs of
displacement
time, velocity-time
and
acceleration-time
for the motion of a
body with constant
acceleration;