The equation for acceleration is \(\displaystyle \frac{\Delta v}{\Delta t}\). The final velocity minus the initial velocity is \(\displaystyle -15 \frac{m}{s}\) and the change in time is \(\displaystyle 5s\). When dividing the change in velocity by the change in time you receive the answer \(\displaystyle -3 \frac{m}{s^{2}}\) which means that the car is decelerating due to the negative sign.

\(\displaystyle \Delta x= (\frac{V_{i}+V_{f}}{2})*t=\frac{9\frac{m}{s}}{2}*3.2s=\mathbf{14.4m}\)

To find the acceleration, use the kinematic equation \(\displaystyle v^2=V_{0}^2+2a(x-x_{0})\)

\(\displaystyle V_{f}=12\frac{m}{s}\)

\(\displaystyle V_{0}=0\)

\(\displaystyle (12\frac{m}{s})^2=2a(50m)\)

\(\displaystyle a=\frac{144\frac{m^2}{s^2}}{100m}=\boldsymbol{1.44\frac{m}{s^2}}\)

*the rider's acceleration will match that of the motorcycle.

Determining horizontal component of velocity:

\(\displaystyle d_x=v_x*t\)

\(\displaystyle v_x=\frac{d_x}{t}\)

\(\displaystyle v_x=\frac{45.7}{5}\)

\(\displaystyle v_x=9.14\frac{m}{s}\)

Using

\(\displaystyle v_x^2+v_y^2=v^2\)

Solving for \(\displaystyle v_y\)

\(\displaystyle v_y=\sqrt{v^2-v_x^2}\)

Combining equations

\(\displaystyle v_y=\sqrt{v^2-(\frac{d_x}{t})^2}\)

Converting \(\displaystyle mph\) to \(\displaystyle mps\)

 \(\displaystyle 50mph*\frac{.447mps}{1 mph}=22.35\frac{m}{s}\)

Converting yards to meters

\(\displaystyle 50yards*\frac{.914 meters}{1 yard}=45.7 meters\)

Plugging in values:

\(\displaystyle v_y=\sqrt{v^2-(\frac{d_x}{t})^2}\)

\(\displaystyle v_y=\sqrt{22.35^2-(\frac{45.7}{5})^2}\)

\(\displaystyle v_y=20.4\frac{m}{s}\)

The ball will be at it's maximum height when the \(\displaystyle y\) component of it's velocity, and thus that direction of kinetic energy, is zero.

The velocity will be decreased by the negative work done by gravity.

\(\displaystyle .5mv_y_i^2+mgh=.5mv_i_f^2\)

Plugging in values:

\(\displaystyle .5*20.4^2+(-9.8)h=.5*0^2\)

Solving for \(\displaystyle h\)

\(\displaystyle h=21.23meters\)

Use the kinematic equation \(\displaystyle V_{f}^2=V_{i}^2+2a(x-x_{0})\).

\(\displaystyle 0=(400\frac{m}{s})^2+2a(0.053m)\)

\(\displaystyle -160,000\frac{m^2}{s^2}=a(0.106m)\)

\(\displaystyle \frac{-160,000\frac{m^2}{s^2}}{0.106m}=a\)

\(\displaystyle \mathbf{a=-1.51*10^6\frac{m}{s^2}}\)

Negative acceleration because the projectile was slowing down.

To find its speed before leaving the ramp,\(\displaystyle V_{f}=V_{0}+at\rightarrow 2.8\frac{m}{s^2}*5s=14\frac{m}{s}\).

Yes, it will be moving \(\displaystyle 2\frac{m}{s}\)faster than necessary. 

Use the kinematic equation \(\displaystyle \Delta x= x_{0}+v_{0}t+.5at^2\).

Since the object was dropped and starts at position 0 in the hand, the first two terms initial position and initial velocity become 0.

\(\displaystyle \Delta x=.5at^2 \rightarrow x=.5(10\frac{m}{s^2})(5)^2\rightarrow \boldsymbol{125 m}\)

Since the object is just falling it is only subject to the acceleration of gravity (especially since we are ignoring frictional forces).

So the final speed is equal to the time multiplied by the acceleration of gravity: \(\displaystyle V_{f}=10 \frac{m}{s^2}*3.1 s= \boldsymbol{31 \frac{m}{s}}\)

The apple is flying straight up with an initial velocity of \(\displaystyle 20\frac{m}{s}\), with the only force acting on it being gravity. Gravity will provide the apple with an acceleration of \(\displaystyle 9.8\frac{m}{s^2}\) directed downwards. The following kinematic equation is needed to solve for the time it takes for the apple to hit the ground:

\(\displaystyle v_f=v_i+a*t\)

The first step is to see how long it takes the apple to get to the highest point of its journey. At that point, the velocity will have slowed to zero:

\(\displaystyle v_i=20\frac{m}{s}\)

\(\displaystyle v_f=0\frac{m}{s}\)

\(\displaystyle a=-9.8\frac{m}{s^2}\)

Solving for t gives us:

\(\displaystyle t=\frac{v_f-v_i}{a}=\frac{0-20}{-9.8}=2.04 s\)

Since it took the apple 2.04 seconds to get to the top, that means it will take the apple another 2.04 seconds to get back to its starting point. This also means that the velocity of the apple when it reaches is starting point will be equal and opposite its starting velocity:

\(\displaystyle v_f=v_i+a*t=0-9.8(2.04)=-20\frac{m}{s}\)

To find the amount of time it takes the apple to reach the ground, the following kinematic equations are used:

\(\displaystyle v_f^2=v_i^2+2ad=-20^2+2*-9.8*-10=596\frac{m^2}{s^2}\)

\(\displaystyle v_f=\sqrt596=24.41\frac{m}{s}\)

And:

\(\displaystyle v_f=v_i+a*t\)

\(\displaystyle t=\frac{v_i-v_f}{a}=\frac{-20+24.41}{9.8}=0.45s\)

Therefore, the total time it takes for the apple to reach the ground is:

\(\displaystyle t_{total}=t_1+t_2+t_3=2.04+2.04+0.45=4.53s\)

We can split this question up into the x-direction and y-direction. In the y-direction, the object's acceleration is zero since it's going to remain in contact with the ground. We now can write our \(\displaystyle \sum F=ma\) equation (let the up direction be positive). Since it's not accelerating, we have

\(\displaystyle \sum F=ma=0\rightarrow F_{n}-mg-3=0\). 

\(\displaystyle F_{n}\) denotes the normal force. We know that the mass is \(\displaystyle 3kg\) and that \(\displaystyle g\approx10\frac{m}{s^2}\), so we can solve for the normal force, which we'll later use to find the force of friction. \(\displaystyle F_{n}-(3kg)(10\frac{m}{s^2})-3=0\)

\(\displaystyle F_{n}=33N\)

The force of friction \(\displaystyle (F_{f})\) is equal to \(\displaystyle \mu F_{n}\) (where \(\displaystyle \mu\) denotes the coefficient of friction), so

\(\displaystyle F_{f}=(.3)(33N)=9.9N\)

In the x-direction, we'll denote right as positive. The forces in the x-direction are \(\displaystyle 40N\) right, \(\displaystyle 3N\) left, and \(\displaystyle 9.9N\). Now we can write our \(\displaystyle \sum F=ma\) equation as \(\displaystyle 40N-3N-9.9N=(3kg)a\)

\(\displaystyle \frac{27.1N}{3kg}=9.0\frac{m}{s^2}\)

Using this acceleration, we can use kinematics to solve for the distance the object will travel in \(\displaystyle 4\) seconds. We have time and acceleration, but we need distance, so we'll use the equation: \(\displaystyle x(t)=(v_{initial})t+0.5at^2\)

\(\displaystyle v_{initial}=0\)

 \(\displaystyle x=0.5(9.0\frac{m}{s^2})(4s)^2=72m\)