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To get the initial velocity and acceleration of the particle... well, we can't. There isn't enough information. I am going to assume that the motion is "smooth:" the final velocity of the initial phase is the same as the constant velocity of the second phase.1. A particle moves for 4 seconds in a straight line with uniform acceleration and describes 52 m. It then travels with uniform speed covering 48 m in 3 s. It is brought to rest by a retardation twice that of the initial acceleration.
So assuming an origin at the point where the particle was when the time started and a +x direction in the direction of the initial acceleration we have:
$\displaystyle v = v_0 + at$
$\displaystyle v^2 = v_0^2 + 2a(x - x_0)$
where $\displaystyle v = \frac{48 \, m}{3 \, s} = 16 \, m/s$
So
$\displaystyle 16 = v_0 + 4a$
$\displaystyle 16^2 = v_0^2 + 2a(52)$
Solving the top equation for v_0 I get:
$\displaystyle v_0 = 16 - 4a$
Inserting this into the second equation gives:
$\displaystyle 256 = (16 - 4a)^2 + 104a$
Solving the quadratic for a I get that:
$\displaystyle a = -5.8503 \, m/s^2$ or $\displaystyle a = 2.3503 \, m/s^2$
We discard the negative solution since the acceleration is defined to be positive in the first phase, so $\displaystyle a = 2.3503 \, m/s^2$.
Thus $\displaystyle v_0 = 16 - 4(2.3503) = 6.59878 \, m/s$
The final retardation (acceleration) is just twice the negative of the acceleration in the first part so $\displaystyle a = -5.0606 \, m/s^2$.
You can use $\displaystyle v^2 = v_0^2 + 2a(x - x_0)$ to find the distance travelled in the last phase of the motion. I leave that to you.
-Dan
Let the starting height of the rock be 0 m (ie. at the origin) and let +y be upward. I will call the height of the ledge h.2. A stone is thrown vertically upward with a velocity of u meters per second. It passes a ledge in $\displaystyle t_1$ seconds and repasses it $\displaystyle t_2$ seconds after the start. Find the height of the ledge.
The stone's acceleration is -g, so the trajectory as a function of time will be
$\displaystyle y = y_0 + ut + \frac{1}{2}at^2$
$\displaystyle y = ut - \frac{1}{2}gt^2$
We know that the stone passes the ledge at two times, thus
$\displaystyle h = ut_1 - \frac{1}{2}gt_1^2$
$\displaystyle h = ut_2 - \frac{1}{2}gt_2^2$
Either of these expressions gives the height of the ledge.
-Dan
By the factor-label method:3. a)Express $\displaystyle 60 km h^{-1}$ in $\displaystyle m s^{-1}$.
$\displaystyle \frac{60 \, km}{1 \, h} \cdot \frac{1000 \, m}{
1 \, km} \cdot \frac{1 \, h}{3600 \, s} = 16.6667 \, m/s$
F = ma3. b) Find the force which would bring a car of mass 800 kg to rest from a speed of 60 km/h in a distance of 50 m.
We need a value for a. Well, we know the change in velocity and the distance that acceleration acted over. So assuming a constant acceleration over this distance we get:
$\displaystyle v^2 = v_0^2 + 2a(x - x_0)$
I will assume an origin where car started to experience the force and a +x direction in the direction of the initial velocity of the car. Thus
$\displaystyle 0^2 = (16.66667)^2 + 2a(50)$
I get $\displaystyle a = 2.7778 \, m/s^2$. Thus
$\displaystyle F = (800 \, kg)(2.7778 \, m/s^2) = 2222.222 \, N$
-Dan