Linear Motion II
Objectives
i) Establish the difference the between acceleration and deceleration
ii) Understand the positive and negative signs of acceleration
iii) Understand and apply the distance formulas in various situations
iv) Familiarize the three equations for linear motion formulas and practice how they are
used in solving questions
Deceleration
- When an object slows down, it is said to be ‘decelerating’, and the sign of the
acceleration will have a negative value
- When an object speeds up, then a>0. When an object slows down, then a<0.
*~ Use a=
for the Ques1) ~Ques3)~*
Q1) A landing airplane makes contact with the runway with a speed of 78m/s. After 18.5
seconds, the airplane comes to a rest. What is the acceleration of the airplane during the
18.5seconds interval?
a) -2.0m/s2
b) -5.3m/s2
c) -2.7m/s2
d) -4.2m/s2
Q2) An object moving at a speed of 10m/s slows down at the rate of 2m/s2. How long
would it take for the object to come to rest?
a) 2 seconds
b) 5 seconds
c) 9 seconds
d) 15 seconds
1
Distance formula for Constant Acceleration
The formula
vf = vi + at ------------------------------------------ (1)
does not allow us to find the distance traveled because the variable ‘d’ is not in the
formula. By using derivatives, from (1) we can obtain
d=(
) -------------------------- (2)
We now have two formulas. If we combine (1) and (2) by eliminating time ‘t’ we can
obtain
(
=
) ----------------------------(3)
This equation is useful when we do not know ‘t’, and not required to find it.
* Formula Summary for Motion with Constant Acceleration
vf = vi + at ------------------------ (1)
use when distance ‘d’ unknown
d= vit + (½)at2 --------------------- (2)
use when final speed ‘vf’ unknown
2ad= vf 2 – vi 2 ---------------------- (3)
use when time ‘t’ unknown
Q3) A truck starts from rest and moves with a constant acceleration of 5m/s2. Find its
speed and the distance traveled after 4s has elapsed. ( Find the distance using formula (2)
& formula (3) and check to see if they give the same result.)
a) 10m/s, 20m
b) 14m/s, 25m
c) 17m/s, 30m
d) 20m/s, 40m
Q4) A car initially at rest accelerates at a constant rate of 4m/s2 until it reaches a speed of
8m/s. How far does the car travel?
a) 4m
b) 8m
c) 18m
d) 25m
2
v= d/ t(constant speed)
a=(vf - vi) / t
d= vit + (½)at2
vf = vi + at
2ad= vf2 – vi2
Q5) A racecar has a speed of 80m/s when the
driver releases a drag parachute. If the
parachute causes a deceleration of -4m/s2, how
far will the car travel before it stops?
a) 20m b) 200m c) 400m d) 800m
Graphic Analysis of ‘distance vs time’ graph for constant acceleration
- The graph for function y=mx2 is shown on right.
Y
- Since distance formula with constant acceleration is
d= vit + (½)at2 and to simply, if the object in motion
started at rest, we can vi =0. Hence, d= (½)at2
y=mx2
X
- The graph for ‘d vs t’ graph for constant
acceleration is shown on right
- As time moves on, the distance will gradually
increase. Hence, the object is speeding up, gaining
more distance as time goes by
- The speed at any instant can be represented as the
slope of the tangent line at any point
- The tangent line becomes steeper corresponding to
time, implying increase in speed.
d
d= (½)at2
t
*~ Snap-shot motion of an object with an acceleration of a=1m/s2 ~*
0m
1m
4m
9m
16m
3
v= d/ t(constant speed)
a=(vf - vi) / t
vf = vi + at
d= vit + (½)at2
2ad= vf2 – vi2
Practice Questions
1. A toy wagon rolls down an inclined plane with constant acceleration. It starts from rest and
attains a speed of 3m/s in 3s. Find the acceleration and the distance moved during that period.
a) 1m/s2 , 4.5m
b) 3m/s2 , 9.0m
c) 5m/s2 , 16m
d) 7m/s2 , 22m
2. A runner is running at a constant speed of 7m/s. The runner gets tired and slows down to a
speed of 2m/s in 5 seconds. What is the acceleration of the runner?
a) 1m/s2
b) -1m/s2
c) 3m/s2
d) -3m/s2
3. A car, starting at rest, accelerates in a straight-line path at a constant rate of 2.5m/s2. How far
will the car travel in 12s?
a) 180m
b) 120m
c) 30m
d) 15m
4. Car-1 is moving at a constant speed of 30m/s. At the instant car-1 pass car-2, car-2 starts from
rest and speeds up to 30m/s in 10 seconds. Which car traveled further in 10 seconds?
5. A motorcycle, starting from rest, has an acceleration of 2m/s 2. What is the speed of the
motorcycle when is moved a distance of 9 meters? + or - ?
a) 13m/s
b) 10m/s
c) 8m/s
d) 6m/s
4
v= d/ t(constant speed)
a=(vf - vi) / t
vf = vi + at
d= vit + (½)at2
2ad= vf2 – vi2
6. The length of the tunnel is 16 meters. A bicycle enters the tunnel at a speed of 10m/s and
leaves the tunnel at 6m/s. How long did it take for the bicycle to go through the tunnel? (hint : use
‘2ad= vf2 – vi2’ to find ‘a’, then use ‘vf = vi + at’ to find the time ‘t’ )
a) 18s
b) 13s
c) 4s
d) 2s
7. A racecar has a speed of 80m/s when the driver releases a drag parachute. If the parachute
causes a deceleration of -4m/s2, how far will the car travel before it stops?
a) 20m
b) 200m
c) 400m
d) 800m
e) 1000m
8. The minimum takeoff speed for a certain airplane is 75m/s. What minimum acceleration is
required if the plane must leave a runway of length 950m? Assume the plane starts from rest at
one end of the runway.
a) 1.50m/s2
b) 2.96m/s2
c) 4.56m/s2
d) 6.43m/s2
9. A car traveling along a road begins accelerating with a constant acceleration of 1.5m/s 2. After
traveling 392m at this acceleration, its speed is 35m/s. Determine the speed of the car when it
began accelerating.
a) 1.5m/s
b) 7.0m/s
c) 34m/s
d) 49m/s
5
v= d/ t(constant speed)
a=(vf - vi) / t
d= vit + (½)at2
vf = vi + at
2ad= vf2 – vi2
10. A rocket-propelled car starts from rest and moves in a straight line with a=5m/s² for 8 seconds
until the fuel is exhausted. It then continues with constant velocity. What distance does the
rocket-car cover in 12 seconds?( *~ That is, 8s when accelerating, then moves 4s more with
constant speed, in total 12s)
a)
b)
d
c)
d
d
t
t
t
10. i) Which graph above represents constant speed?
ii) Which graph above represents decreasing speed?
iii) Which graph above represents increasing speed?
11. A cheetah is walking at a speed of 1.1m/s when it observes a rabbit 41m directly ahead. If the
cheetah accelerates at 9.55m/s2, how long does it take the cheetah to reach the rabbit if the rabbit
doesn’t move?
( you may have to use the quadratic formula; when ax2 + bx +c =0, then x =
a) 8.45s
b) 5.67s
c) 3.44s
6
)
d) 2.82s