A car and a bus are side by side at the traffic lights.
They are both stationary. The lights turn to green.Thirty seconds later the bus and the car are at
the same speed of 15m/s (approx 30 mph)
The car however is ahead of the bus.
a)
Does
the car have a higher speed than the bus?
b)
How
much does the car’s speed go up?
c)
How
much does the bus’ speed go up
d)
Which
took less time to reach 15 m/s the car or the bus?
e)
Which
has the higher acceleration the car or the bus?
f)
Does
acceleration just depend upon changing speed? (More than yes or no is required
here)
Calculations
1.
Jerry is standing in the middle of the room Tom
runs around the corner. Jerry accelerates to 5m/s in 3 s. Calculate the value
of his acceleration.
a = (v-u)/t =
(5-0)/3 = 1.67 ms-2
2.
Tom is just behind Jerry when he runs into a
mouse hole. Tom hits the wall at a speed of 6m/s. He comes to rest in 0.2s.
Calculate the acceleration of Tom’s head. Does the collision kill Tom?
a = (v-u)/t =
(0-6)/0.2 = -30 ms-2 No,
cartoon characters follow the laws of “Wile E Coyote Physics”
3.
At the end of a 215km race Mark accelerates from
18m/s to 21 m/s in 1.2s. Calculate his acceleration.
a = (v-u)/t = (21-18)/1.2
= 2.5 ms-2
4.
Jeremy pushes the pedal to the metal once again.
His reasonably priced car takes 13.6 seconds to accelerate from 13m/s to 47m/s.
Calculate his acceleration.
a = (v-u)/t =
(47-13)/13.6 = 2.5 ms-2
5.
Thomas is puffing along at 25m/s when the signal
ahead turns to red. He applies his brakes and slows to 11m/s in 35s. Calculate
his acceleration.
a = (v-u)/t =
(11-25)/35 = -0.4 ms-2
6.
An oil tanker’s top speed is 6.7 m/s. If it
decelerates at 0.005m/s2. Calculate how long it takes to stop.
a = (v-u)/t
rearranging t =
(v-u)/a
t = (0 – 6.7)/ -
0.005
t = 1340s = 22 min
Distance = t(v-u)/2
= 1340(0-6.7)/2 = 4489m = 4.5km
7.
At take off the Lunar Module had an acceleration
of 3.4 m/s2. The moon’s gravitational field will accelerate objects
at 1.8m/s2. Calculate the velocity of the Eagle 3s after lift off
from Tranquillity Base.
a = (v-u)/t,
rearranging (v-u) =
at,
(v-0) = (3.4 – 1.8)
3
v = 4.8 ms-1
8.
At 11.40pm on the 14th April 1912 the
Titanic was running at 22 knots (11.3 m/s) when it hit an iceberg. It is
popular lore that the iceberg was spotted 37 seconds before the ship hit it. After
the disaster tests were performed on its sister ship the Olympic. At 18 knots
(9.2 m/s) it took 3 min 15 seconds to stop. Calculate the deceleration of The
Olympic.
a = (v-u)/t a= (0-9.2)/(3 x 60) +15, a = -9.2/195 = -0.05 ms-2
Assume the Titanic decelerated at the same
rate, and the officers on the bridge had reacted immediately. Calculate it’s
velocity after 37 seconds of deceleration.
a = (v-u)/t,
rearranging (v-u)= at,
(v – 11.3) = -0.05 x 37,
(v – 11.3) = -1.74,
v = 11.3 – 1.74 =
9.6 ms-2
(The 37 seconds was calculated after the
event. It was the time needed for the Titanic to have swerved away from the
iceberg)
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