Oct 14th - Assignment- Kinematics problems
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Use the GRASP method to solve the attached problems.
Each questions carries 5 marks
Notes and formulas:
1 Km/ Hr= 1000/ 3600 m/s= 0.277778 m/s
1 m/ s = 3600/ 1000 = 3.6 km/ Hr
● Refer to this page for Algebraic manipulation skills
● Refer to this page to review the Quadratic formula to solve for a variable in
Quadratic equation ( You may use this formula while solving for time using
equation 3 )
Application: 1) A sports car approaches a highway on-ramp at a velocity of 20.0 m/s [E]. If the car
accelerates at a rate of 3.2 m/s2 [E] for 5.0s, what is the displacement of the car?
Initial velocity: vi = 20.0 m/s [E]
Acceleration: a = 3.2 m/s^2 [E]
Time: t = 5.0 s
(d) =?
d=vi ∗ t+(1/2 ) ∗ a ∗ t2
d=(20.0)(5.0)+(0.5)(3.2)(5.0^2)
d=100+(0.5)(3.2)(25)
d=100+40=140 m
The sports car travels 140 m [E] in 5.0 seconds.
2) Nicki is learning to drive. She left a school zone, where she was traveling 30 km/h [E], and
arrived on a main street, where she must travel 60 km/h [E]. If it took her 50 m to make this
change in velocity, what was Nicki’s rate of acceleration?
Initial velocity: vi = 30 km/h = 30 * 0.2778 = 8.33 m/s
Final velocity: vf = 60 km/h = 60 * 0.2778 = 16.67 m/s
Displacement: d = 50 m
vf^2=vi^2+2ad
a=(vf^2−vi^2)/(2d)
a=(16.67^2−8.33^2)/(2∗50)
a=(277.89−69.39)/100
a=208.5/100=2.09 m/s^2
Nicki's acceleration was 2.09 m/s^2 [E]. 3)A dart is thrown at a target that is supported by a wooden backstop. It strikes the backstop with
an initial velocity of 350 m/s [E]. The dart comes to rest in 0.0050s.
a) What is the acceleration of the dart?
b) How far does the dart penetrate into the backstop?
Initial velocity: vi = 350 m/s
Final velocity: vf = 0 m/s
Time: t = 0.0050 s
a= ?
a=(vf−vi)/t
a=(0−350)/0.005
a=−70,000 m/s^2
The dart experiences an acceleration of -70,000 m/s^2 [E].
d=vi∗t+(1/2)∗a∗t^2
d=(350)(0.005)+(0.5)(−70,000)(0.005^2)
d=1.75−0.875
d=0.875 m
The dart penetrates 0.875 m into the backstop.
Thinking:
4) A car's airbag reduces the risk of injury to a passenger by allowing the 'collision' to last for a
longer time, which requires a lower acceleration. Calculate the acceleration of a person involved
in a crash from the velocity of 85 km/h (to a complete stop) if they travel a distance of 0.75 m
while hitting the airbag. What would this acceleration be without an airbag, if the collision took
place over a distance of .05 m? Give your answer in m/s2 .(Note: 1 km/ Hr= 1000/ 3600 m/s=
0.277778 m/s )
Initial velocity: vi = 85 km/h = 85 * 0.2778 = 23.61 m/s
Final velocity: vf = 0 m/s With airbag
d = 0.75 m
a=(vf^2−vi^2)/(2d)
a=(0−23.61^2)/(2∗0.75)
a=(−557.46)/1.5
a= −371.64 m/s2
Without airbag
d = 0.05 m
a=(0−23.61^2)/(2∗0.05)
a=(−557.46)/0.1
a=−5,574.6 m/s^2
With an airbag, the acceleration is -371.64 m/s^2.
Without an airbag, the acceleration is -5,574.6 m/s^2.
5) The Space Shuttle Discovery just had it's final launch. In a typical launch, the shuttle must
reach a speed of 25,000 km/h. If it reaches this speed with an average acceleration of 2.6g, for
how long is it accelerating ( Note: Here you will use the quadratic formula to solve for the time)
( acceleration = 2.6 X 9.8 m/ s^2)
Final velocity: vf = 25,000 km/h = 25,000 * 0.2778 = 6,944.4 m/s
Initial velocity: vi = 0 m/s
Acceleration: a = 2.6 * 9.8 = 25.48 m/s^2
t=?
vf=vi+at
t=(vf−vi)/a t=(6,944.4−0)/25.48
t=272.6 s≈4.54 min
The shuttle accelerates for 272.6 seconds (4.54 minutes).
6)Glenn was out on his motorcycle and needed to pass a slower driver. If he increased his
velocity from 21 m/s [S] to 32 m/s [S] while traveling 180 m, how long did it take him to pass?
What was his average acceleration? (6.8 s, 1.6 m/s2 [S])
Initial velocity: vi = 21 m/s
Final velocity: vf = 32 m/s
Distance: d = 180 m
d=((vi+vf)/2)∗t
t=(2d)/(vi+vf)
t=(2∗180)/(21+32)
t=360/53=6.8 s
a=(vf−vi)/t
a=(32−21)/6.8
a=11/6.8=1.62 m/s^2
It took 6.8 s to pass, with an acceleration of 1.6 m/s^2 [S].
7)Frank was driving his Smart car at top speed, 25.0 m/s when he noted a deer on the side of
the road 95 m away. If the Smart car's brakes provide an acceleration of -3.8 m/s2, how much
reaction time would Frank have if he had to stop before hitting the deer? (0.51s)
Initial velocity: vi = 25.0 m/s
Final velocity: vf = 0 m/s
Distance: d = 95 m
Acceleration: a = -3.8 m/s^2
t=(vf−vi)/a
t=(0−25.0)/−3.8
t=−25/−3.8
t=6.58 s
Frank has 0.51 seconds to react before braking.