Constant Acceleration Problems
1. The physics of an accelerating spacecraft.
A spacecraft is moving at a speed of +3550 m/s. The retrorockets of the spacecraft
are then fired and the spacecraft begins to slow down with an acceleration whose
magnitude is 10.0 m/s2. Assume that the displacement of the craft is +225 km.
What is the velocity of the craft after this displacement?
2. The physics of an accelerating skier.
a) What is the magnitude of the average acceleration of a skier who, starting from
rest, reaches a velocity of -8.0 m/s when going down a slope for 4.0 s?
b) How far does the skier travel in this time?
3. The physics of an accelerating rocket.
During an interval of 20.0 s, a rocket's velocity increased from 255 m/s to 555
m/s. What was the displacement of the rocket during this time interval?
Solutions
1. The physics of an accelerating spacecraft
A spacecraft is moving at a speed of +3550 m/s. The retrorockets of the spacecraft
are then fired and the spacecraft begins to slow down with an acceleration whose
magnitude is 10.0 m/s2. Assume that the displacement of the craft is +225 km.
What is the velocity of the craft after this displacement?
What do we know?
v2 = ?
v1 = 3550m/s
a=-10.0 m/s2
∆d=225km
Select the appropriate equation and solve the problem.
The appropriate equation for this question would be v22 = v12 +2ad
v22 = 35502 +2x-10x225
v2 = √(35502 +2x-10x225)
= +2850 m/s or - 2850 m/s.
Both answers are possible. In the case where v2 = +2850 m/s, the spacecraft is
still moving to the right after it has slowed down. This corresponds to the
diagram above. In the case where v2 = -2850 m/s, the spacecraft has now
reversed direction and is moving to the left. This second situation would take
more time than the first.
2.
The physics of an accelerating skier.
a) What is the magnitude of the average acceleration of a skier who, starting from
rest, reaches a velocity of -8.0 m/s when going down a slope for 4.0 s?
b) How far does the skier travel in this time?
Decide which direction is positive and which is negative.
It is convenient to let the direction up be positive and the direction down to
be negative.
v1 = 0m/s
v2 = -8m/s
∆t = 4s
a=? ∆d = ?
Select the appropriate equation and solve the problem.
a) a = (v2 – v1) = -8 – 0 = -2m/s2
∆t
4
∆d = v1∆t + 2a∆t2 = 0x4 + (-2)x(4)2 = b) Calculating distance travelled. The appropriate equation for this question
is ∆d = v1∆t + 2a∆t2 = (0.0 m/s)(4.0 s) + (-2.0 m/s2)(4.0 s) 2 = -16. M
3. The physics of an accelerating rocket.
During an interval of 20.0 s, a rocket's velocity increased from 255 m/s to
555 m/s. What was the displacement of the rocket during this time interval?
Decide which direction is positive and which is negative.
It is convenient to let the direction to the right be positive and the direction to the
left to be negative.
We know
v1 = 255m/s
v2 = 555m/s
∆t = 20s
∆d = ?.
d = (v1+v2) ∆t = (255 m/s + 555 m/s)(20.0 s) = 8100 m.
2
2
Relative Motion Problems
1. The physics of a hound walking on a moving boat.
A hound walks at a speed of 2.0 m/s along the deck toward the front of a boat
which is travelling at 8.0 m/s with respect to the water.
a) What is the velocity of the hound relative to the water?
b) What would be the velocity of the hound if the dog were walking toward the
back of the boat?
2. The physics of an airplane of an airplane travelling in calm air.
An Air Canada plane is travelling at 1000.0 km/h in a direction 40.00° east of
north.
a) Find the components of the velocity vector by finding the component in the
northerly direction and the easterly direction.
b) How far to the north and how far to the east would the plane travel in 4.00 h?
3. The physics of a vacationer and a cruise ship.
A captain walks 4.0 km/h directly across a cruise ship whose speed relative to the
earth is 12.0 km/h. What is the speed of the captain with respect to the earth.
4. The physics of a swimmer in a current.
A swimmer is capable of swimming at a speed of 1.5 m/s in still water. This is
equivalent to saying that the swimmer can swim with a speed of 1.5 m/s relative
to the water. The swimmer starts to swim directly across a 3.0 km wide river.
However, the current is 0.90 m/s, and it carries the swimmer downstream.
a) How long does it take the swimmer to cross the river?
b) How far downstream will the swimmer be upon reaching the other side of the
river?
Solutions
1a) Let us assume for the sake of simplicity, that the boat is moving to the
right.
The velocity of the boat with respect to the water is vbw.
The velocity of the hound with respect to the boat is vhb.
The velocity of the hound with respect to the water is vhw.
vhw = vhb + vbw = 2.0 + 8.0 = 10.0 m/s
b) In this case, the velocity of the dog is opposite to that of the boat.
vhw = vhb + vbw = -2.0 + 8.0 = 6.0 m/s.
2. The physics of an airplane of an airplane travelling in calm air.
An Air Canada plane is travelling at 1000.0 km/h in a direction 40.00° east of north.
a) Find the components of the velocity vector by finding the component in the
northerly direction and the easterly direction.
b) How far to the north and how far to the east would the plane travel in 4.00 h?
A rough diagram of the motion will help to
visualize the situation.
a) The component in the northerly direction is
1000.0 (cos 40.00°) = 766.0 km/h.
The component in the easterly direction is
1000.0 (sin 37.00°) = 642.8 km/h.
b) To find the distances travelled, use d = vt.
The distance travelled in the northerly direction is
d = (766.0)(4.00) = 3.06 x 103 km.
The distance travelled in the easterly direction is
d = (642.8)(4.00) = 2.57 x 103 km.
3 The physics of a vacationer and a cruise ship.
A captain walks 4.0 km/h directly across a cruise ship whose speed relative to the earth is
12.0 km/h. What is the speed of the captain with respect to the earth.
In this case, the velocity of the captain with respect to the earth is the
hypotenuse of the triangle.
vCS is the velocity of the captain with respect to the ship.
vSE is the velocity of the ship with respect to the earth.
vCE is the velocity of the captain with respect to the earth.
vCE is 12 2  4 2 = 12.6 km/h.
Note that the angle is not stated since the question asked for speed and not
velocity.
4. The physics of a swimmer in a current.
A swimmer is capable of swimming at a speed of 1.5 m/s in still water. This is
equivalent to saying that the swimmer can swim with a speed of 1.5 m/s relative
to the water. The swimmer starts to swim directly across a 3.0 km wide river.
However, the current is 0.90 m/s, and it carries the swimmer downstream.
a) How long does it take the swimmer to cross the river?
b) How far downstream will the swimmer be upon reaching the other side of the
river?
The velocity vSG of the swimmer relative to the ground is the vector sum of
the velocity vSW of the swimmer relative to the water and the velocity vWG of
the water relative to the ground as shown below:
vSG = vSW + vWG.
a) The component of vSG that is parallel to the width of the river determines
how fast the swimmer is moving across the river; this parallel component is
vSW. The time for the swimmer to cross the river is equal to the width of the
river divided by the magnitude of this velocity component.
t = width / vSW = 3.0 x 103 / 1.5 = 2.0 x 103 s.
b) The component of vSG that is parallel to the direction of the current
determines how far the swimmer is carried down stream; this component is
vWG. Since the motion occurs with constant velocity, the distance that the
swimmer is carried downstream while crossing the river is equal to the
magnitude of vWG multiplied by the time it takes for the swimmer to cross the
river.
x = vWGt = (0.90)(2.0 x 103) = 1.8 x 103 m.