A particle has an initial horizontal velocity
of 2.9 m/s and an initial upward velocity of
4.5 m/s. It is then given a horizontal accelera-
tion of 1.2 m/s2 and a downward acceleration
of 1.2 m/s2.
What is its speed after 4.9 s?
Answer in units of m/s
What is the direction of its velocity at this
time with respect to the horizontal? Answer
between −180� and +180�.
Answer in units of �
To find the speed of the particle after 4.9 s, we can use the equations of motion.
Given:
Initial horizontal velocity (u_x) = 2.9 m/s
Initial upward velocity (u_y) = 4.5 m/s
Horizontal acceleration (a_x) = 1.2 m/s^2
Downward acceleration (a_y) = 1.2 m/s^2
First, we need to find the final velocities in horizontal and vertical directions after 4.9 s.
Horizontal velocity (v_x) after 4.9 s:
We can use the equation:
v_x = u_x + a_x * t
where:
v_x - final horizontal velocity
u_x - initial horizontal velocity
a_x - horizontal acceleration
t - time
Substituting the known values:
v_x = 2.9 m/s + (1.2 m/s^2) * (4.9 s)
v_x = 2.9 m/s + 5.88 m/s
v_x = 8.78 m/s
Vertical velocity (v_y) after 4.9 s:
We can use the equation:
v_y = u_y + a_y * t
where:
v_y - final vertical velocity
u_y - initial upward velocity
a_y - downward acceleration
t - time
Substituting the known values:
v_y = 4.5 m/s + (-1.2 m/s^2) * (4.9 s)
v_y = 4.5 m/s - 5.88 m/s
v_y = -1.38 m/s
Now, let's find the total speed (v) after 4.9 s:
The total speed is the magnitude of the velocity vector, which can be found using the Pythagorean theorem:
v = sqrt(v_x^2 + v_y^2)
Substituting the calculated values:
v = sqrt((8.78 m/s)^2 + (-1.38 m/s)^2)
v = sqrt(77.1684 m^2/s^2 + 1.9044 m^2/s^2)
v = sqrt(79.0728 m^2/s^2)
v ≈ 8.8 m/s
Therefore, the speed of the particle after 4.9 s is approximately 8.8 m/s.