Electrical current is defined as
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Answer:
i) The motion of both particles are shown on the same speed-time curve included
ii) Approximately 19.5 seconds
Explanation:
We are given that;
Initial velocity of particle, P = 140 m/s
Start time of particle P = 6 s before start time of particle Q
Position of particle Q when velocity is 25 m/s = 125 m
Therefore, from the equation of motion, we have for particle Q;
v² = u² + 2·a·s
Where:
v = Final velocity = 25 m/s
u = Initial velocity = 0 m/s
a = Acceleration
s = Distance covered = 125 m
Therefore;
25² = 0² + 2×a×125
Which gives a = 25²/(2×125) = 2.5 m/s²
The time taken for particle Q to reach 125 m is found from the relation;
s = u·t + 1/2·a·t²
Where:
t = Time of journey
Therefore;
125 = 0×t + 1/2×2.5×t²
Which gives 125 = 1.25 × t²
Hence, t² = 125/1.25 = 100
t = √(100) = 10 s
The equation for particle Q is v = 0 + 2.5×t
Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;
Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s
From the equation of motion, for particle P (decelerating) we have
v = u - a·t
Where:
v = 25 m/s
u = 140 m/s
t = 16 s
Hence, 25 = 140 - a×16
∴ 16·a = 140 - 25 = 115
a = 115/16 = 7.1875 m/s²
Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;
v = u - a·t
0 = 140 - 7.1875 × t
∴7.1875·t = 140
t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.
1) -0.5 m/s
We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.
Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:
After the shot, the momentum is:
where
M = 2000 kg is the mass of the cannon
m = 10 kg is the mass of the shell
v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)
V = ? is the velocity of the cannon
Since momentum is conserved, we can write
And solving for V, we find the velocity of the cannon:
where the negative sign indicates that the cannon moves in the direction opposite to the shell.
2) 0.5 m
The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:
where
v is the final velocity of the cannon
u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)
is the deceleration of the cannon
s is the distance travelled by the cannon
The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon: