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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:
where
So
Solving the top equation for v_0 I get:
Inserting this into the second equation gives:
Solving the quadratic for a I get that:
or
We discard the negative solution since the acceleration is defined to be positive in the first phase, so .
Thus
The final retardation (acceleration) is just twice the negative of the acceleration in the first part so .
You can use 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 seconds and repasses it 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
We know that the stone passes the ledge at two times, thus
Either of these expressions gives the height of the ledge.
-Dan
By the factor-label method:3. a)Express in .
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:
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
I get . Thus
-Dan