A well-labeled sketch usually helps make the problem clearer.

(a) Next, we list the list the given information and what we are looking for:

v0 = +5.0 m/s

vf = -1.5 m/s

Δx = +5.5 m

a = ?

vaverage = ?

t = ?

Note that I have taken the direction up the incline as positive and that the signs are explicitly stated. It is a very common source of error to leave out or to not consider the signs of directions of all vector quantities.

To find the acceleration, we find the kinematics equation that contains a and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find a yields . Notice that the acceleration is negative. This means that the acceleration points down the incline. It means that an object traveling up an incline will slow, turn around, and roll down the incline.

The average velocity is defined .

To find the time, we find the kinematics equation that contains a and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find t yields .

(b) When an object moving in 1D turns around we know that the object is instantaneously at rest and that its velocity at that point is v₃ = 0. The information that we know is thus:

v0 = +5.0 m/s
v3 = 0 m/s	This is our new final velocity
a = -2.068 m/s2	From part (a)
Δx = ?
vaverage = ?
t = ?

Notice that the acceleration is a constant of the motion; it has the same value in both parts of the problem.

To find the displacement from the initial position where the ball turns around, we find the kinematics equation that contains x and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find x yields . Notice that this value is bigger than the original 5.5 m and is consistent with the sketch, i.e. the ball was farther up the incline when it turned around.

To find the time it takes for the ball to reach the point where it turns around, we find the kinematics equation that contains t and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find t yields .Notice that this value is smaller than the time in part (a) and is consistent with the sketch, i.e. the ball hasn't come back down the incline yet.

To solve this problem, we list the given information and what we are looking for:

v0 = 0.0 m/s	since the bullet is initially at rest
vf = 500 m/s	velocity of the bullet as it leaves the barrel
a = 70,000 m/s
Δx = ?	the length of the barrel
t = ?	the time it takes to travel the barrel
vaverage = ?

To find the length of the barrel, we find the kinematics equation that contains x and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find a yields .

To find the time it takes for the bullet to travel the length of barrel, we find the kinematics equation that contains t and the given quantities. Examining our equations we see that we can use . Rearranging this equation to find t yields .

The average velocity is defined .

To solve this problem, we list the list the given information

Red Car	Blue Car
v0 red = 0.0 m/s	v0 blue = 62.0 km/h = 17.222 m/s
ared = 2.00 m/s2	ablue = 0 m/s2 (constant velocity)
Δxred = ?	Δxblue = ?
tred = ?	tblue = ?

This is an example of a two-body constrained kinematics problem. Even if a sketch was not explicitly required, we would need one anyway to get the constraints. For the sketch, recall that on a d versus t curve an object moving forward with a uniform acceleration should be represented by a line curving upwards while an object with constant forward velocity is represented by a straight line with a positive slope.

Looking at the sketch, we see that our constraints are:

To solve the problem, we must find the kinematics equation that contains the known quantities, v₀ and a, and the unknown quantities, Δx and t. Examining our equations we see that we can use Δx= v₀t + ½at². We substitute this equation into both sides of equation (1). This yields,

We then use equation (2) to replace t_red and t_blue by t,

Plugging in the values of the given quantities yields,

The solution of this equation is t = 17.222 seconds. This is the time that elapses before the two cars meet again.

With a value for t, we can find how far down the road the red car has traveled;

As a check, we can find how far down the road the blue car has traveled;

So the cars meet 297 m down the road.

According to our definition of average velocity, v_{average red}= Δx_red/t = (297 m)/(17.2 s) = 17.2 m/s. Since the blue car maintains a constant velocity, v_{average blue}= v_{0 blue} = 17.2 m/s. The two quantities are the same since the two cars have traveled the same distance in the same amount of time.

To solve this problem, we list the list the given information

Constable	Motorist
v0 police = 0.0 m/s	v0 speeder = 110 km/h = 30.556 m/s
apolice = 2.00 m/s2	aspeeder = 0 m/s2 (constant velocity)
Δxpolice = ?	Δxspeeder = ?
tpolice = ?	tspeeder = ?

This is an example of a two-body constrained kinematics problem. We need a sketch to get the constraints. For the sketch, recall that on a d versus t curve an object moving forward with a uniform acceleration should be represented by a line curving upwards while an object with constant forward velocity is represented by a straight line with a positive slope.

Looking at the sketch, we see that our constraints are:

To solve the problem, we must find the kinematics equation that contains the known quantities, v₀ and a, and the unknown quantities, Δx and t. Examining our equations we see that we can use Δx= v₀t + ½at². We substitute this equation into both sides of equation (1). This yields,

We then use equation (2) to replace t_speeder by t_police + 1,

Plugging in the values of the given quantities yields,

This is a quadratic in t_police. Solving the quadratic yields, t_police = 30.07 seconds. It takes the police constable 30.1s to catch up with the speeder. The speeder was traveling for 31.1 s.

With a value for t_police, we can find how far down the road the police car has traveled;

As a check, we can find how far down the road the speeder's car has traveled;

So the cars meet 949 m down the road.

To solve this problem, we list the list the given information and what we are looking for:

v0 = 12.0 m/s	velocity as it leaves the hand
vtop = 0 m/s	since it turns around
vf = -12.0 m/s	symmetry says it must have this value when it returns to the same height
a = -9.81 m/s	only gravity is acting
Δy = 0	since it returns to the same height
tair = ?	the time it takes for the entire trip
tup = tdown = ½tair = ?	symmetry requires this

We have lots and lots of information from symmetry. To find t_air, choose the kinematics equation that has t and the known quantities v₀, v_f, and a, that is v_f = v₀+ at_air. Solving yields t_air = (v_f - v₀)/a = (-v₀-v₀)/(-g) = 2v₀/g = 2.4465 seconds. Hence t_up = t_down = 1.2232 s.

To find h, choose the kinematics equation that has Δy (h is a displacement) and the known quantities v₀, v_top, and a, that is . Upon rearrangement, this yields h = Δy = (v₀)²/g = 7.34 m.

To solve this problem, we list the list the given information and what we are looking for:

v0 = ?	velocity as it leaves the hand
vtop = 0 m/s	since it turns around
vf = -v0	symmetry says it must have this value when it returns to the same height
a = -9.81 m/s	only gravity is acting
Δy = 0	since it returns to the same height
tair = 3.20 s	the time it takes for the entire trip
tup = tdown = ½tair = 1.60 s	symmetry requires this

We have lots and lots of information from symmetry. To find v₀, choose the kinematics equation that has v₀ and the known quantities, v_f = -v₀ , t_air and a, that is v_f = v₀+ at_air. Eliminating v_f yields -v₀ = v₀ - gt_air. Rearranging gives v₀ = gt_air/2 = 15.7 m/s.

To find h, choose the kinematics equation that has Δy (h is a displacement) and the known quantities v₀, v_top, and a, that is . Upon rearrangement, this yields h = Δy = (v₀)²/g = 12.6 m.

To solve this problem, we list the list the given information

Ball #1	Ball #2
v0 1 = 15.0 m/s	v0 2 = 12.0 m/s
a1 = -9.81 m/s2	a2 = -9.81 m/s2
Δy1 = ?	Δy2 = ?
t1 = ?	t2 = ?

This is an example of a two-body constrained kinematics problem. We need a sketch to get the constraints. For the sketch, recall the shape of the d versus t curve for an object thrown up into the air - a parabola.

Looking at the sketch, we see that our constraints are:

To solve the problem, we must find the kinematics equation that contains the known quantities, v₀ and a = -g, and the unknown quantities, Δy and t. Examining our equations we see that we can use Δy = v₀t - ½gt². We substitute this equation into both sides of equation (1). This yields,

We then use equation (2) to replace t₁ by t₂ + 1.15,

This reduces to

Upon rearrangement this becomes

Thus t₂ = 1.2997 s, and t₁ = 2.4497 s. Now that we have the time that each ball is in the air, we can now find h

So the balls collide when they are 7.31 m in the air.