horizontal:
2.1*7.9= ...
vertical:
5*7.9=....
speed= sqrt(horizontal^2+ vertical^2)
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.
2.1*7.9= ...
vertical:
5*7.9=....
speed= sqrt(horizontal^2+ vertical^2)
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!
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.