A particle at rest undergoes an acceleration
of 2.1 m/s2 to the right and 5 m/s2 up.
a) What is its speed after 7.9 s?
Answer in units of m/s.
horizontal:
2.1*7.9= ...
vertical:
5*7.9=....
speed= sqrt(horizontal^2+ vertical^2)
Well, if a particle can accelerate even when it's at rest, it must have a very active imagination! Maybe it dreams of becoming a track star one day. Anyway, to find the speed after 7.9 seconds, we can use a combination of the horizontal and vertical accelerations.
First, let's find the horizontal component of the speed. The particle experiences an acceleration of 2.1 m/s^2 to the right, so after 7.9 seconds, its horizontal speed will be:
Horizontal speed = initial horizontal speed + horizontal acceleration * time
Since the particle was initially at rest, the initial horizontal speed is 0. So we're left with:
Horizontal speed = 2.1 m/s^2 * 7.9 s
Now, let's calculate the vertical component of the speed. The particle undergoes an acceleration of 5 m/s^2 up, so after 7.9 seconds, its vertical speed will be:
Vertical speed = initial vertical speed + vertical acceleration * time
Again, since the particle was initially at rest, the initial vertical speed is 0. So we have:
Vertical speed = 5 m/s^2 * 7.9 s
Now, we need to combine the horizontal and vertical speeds to find the overall speed. We can use the Pythagorean theorem to do that:
Speed = sqrt((Horizontal speed)^2 + (Vertical speed)^2)
And there you have it - the speed of our speedy particle after 7.9 seconds!
To find the speed of the particle after 7.9 seconds, we can use the concept of vector addition to combine the horizontal and vertical components of acceleration.
First, let's break down the problem into its horizontal and vertical components:
Horizontal component:
Acceleration in the horizontal direction = 2.1 m/s²
Time, t = 7.9 s
Using the formula for uniform acceleration:
v = u + at
where:
v = final velocity
u = initial velocity (which is zero since the particle is at rest)
a = acceleration
t = time
Substituting the values:
v_horizontal = 0 + (2.1 m/s²)(7.9 s)
v_horizontal = 16.59 m/s
Vertical component:
Acceleration in the vertical direction = 5 m/s²
Time, t = 7.9 s
Using the formula for uniform acceleration:
v = u + at
Substituting the values:
v_vertical = 0 + (5 m/s²)(7.9 s)
v_vertical = 39.5 m/s
Now, we can find the magnitude of the total velocity (speed) of the particle using the Pythagorean theorem:
speed = √(v_horizontal² + v_vertical²)
speed = √((16.59 m/s)² + (39.5 m/s)²)
Calculating this expression gives us the final answer:
speed = √(275.0081 m²/s² + 1560.25 m²/s²)
speed = √(1835.2581 m²/s²)
speed ≈ 42.85 m/s
Therefore, the speed of the particle after 7.9 seconds is approximately 42.85 m/s.