Two particles moving along parallel paths in
the same direction pass the same point at the
same time. Particle A has an initial velocity
of 8.7 m/s and an acceleration of 1.2 m/s^2
Particle B has an initial velocity of 4.3 m/s
and an acceleration of 7.7 m/s^2
At what time will B pass A?
Answer in units of s
At what position will the faster particle
pass the slower?
Answer in units of m
5rytfty
To find the time at which particle B passes particle A, we can use the equations of motion and solve for time (t).
For Particle A:
Initial velocity (u) = 8.7 m/s
Acceleration (a) = 1.2 m/s^2
For Particle B:
Initial velocity (u) = 4.3 m/s
Acceleration (a) = 7.7 m/s^2
Let's start with finding the time at which B passes A:
We can use the equation:
s = ut + (1/2)at^2
For Particle A:
Let's take the starting position of Particle A as the reference point, so s = 0.
0 = (8.7 × t) + (0.5 × 1.2 × t^2)
0 = 8.7t + 0.6t^2 ---(equation 1)
For Particle B:
We need to calculate the position of Particle B in terms of time.
s = (4.3 × t) + (0.5 × 7.7 × t^2)
s = 4.3t + 3.85t^2 ---(equation 2)
We need to find the time (t) at which both particles pass the same point. In other words, we need to find the value of t where s is equal for both particles.
Setting equation 1 equal to equation 2:
8.7t + 0.6t^2 = 4.3t + 3.85t^2
Rearranging and simplifying:
3.25t^2 - 4.4t = 0
t(3.25t - 4.4) = 0
This equation has two solutions: t = 0 and t = 1.35
Since both particles pass the point at the same time, the answer is t = 1.35 seconds.
Now, let's find the position at which the faster particle (Particle B) passes the slower particle (Particle A).
We can substitute the value of t (1.35 seconds) into the equation for Particle B's position (equation 2):
s = 4.3(1.35) + 3.85(1.35)^2
s = 5.805 + 6.205
s = 12.01 meters
So, Particle B passes Particle A at a position of 12.01 meters.