A baseball is hit straight up into the air at an initial velocity of 35 m/s. How long does it take for the
baseball to reach its highest position?
and
As a traffic light turns green, a car at rest begins to move forward at a uniform acceleration of 3.1 m/s^2 [E] the same time, a truck passes through the intersection travelling at a constant velocity of 15 m.s [E] At what time does the car pass the truck?
v = 35 - 9.81t
when is v=0?
1/2 * 3.1 t^2 = 15t
To find the time it takes for the baseball to reach its highest position, we can use the kinematic equation for motion with constant acceleration:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity is 35 m/s, and the final velocity when the ball reaches its highest position is 0 m/s (since it momentarily stops before falling back down). The acceleration due to gravity acting on the ball is approximately -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity).
Using these values, we can rearrange the equation to solve for time (t):
0 = 35 + (-9.8)t
Simplifying the equation gives:
-9.8t = -35
Dividing both sides by -9.8:
t = 35 / 9.8
Calculating this gives:
t ≈ 3.57 seconds
Therefore, it takes approximately 3.57 seconds for the baseball to reach its highest position.
Now let's move on to the second question about the car passing the truck.
To find the time when the car passes the truck, we need to calculate the time it takes for the car to accelerate to the same velocity as the truck.
The formula for velocity (v) using acceleration (a), initial velocity (u), and time (t) is:
v = u + at
In this case, the initial velocity of the car is 0 m/s since it starts from rest, and the acceleration is 3.1 m/s^2. The initial velocity of the truck is 15 m/s, and it travels at a constant velocity.
Setting the velocities of the car and the truck equal to each other, we have:
0 + 3.1t = 15
Simplifying the equation gives:
t = 15 / 3.1
Calculating this gives:
t ≈ 4.84 seconds
Therefore, the car passes the truck at approximately 4.84 seconds after the traffic light turns green.
To find the time it takes for the baseball to reach its highest position, we can use the kinematic equation for vertical motion:
h = v0*t - (1/2)*g*t^2
Where:
h = height of the baseball
v0 = initial velocity of the baseball (35 m/s)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
At the highest position, the baseball's final velocity will be 0 m/s. So we can set the final velocity (v) equal to 0 in the equation:
0 = v0 - g*t
Solving for t gives us:
t = v0 / g
Substituting the given values:
t = 35 m/s / 9.8 m/s^2
Calculating this value will give us the time it takes for the baseball to reach its highest position.
For the second question, we can determine the time at which the car passes the truck by first finding the time it takes for the car to accelerate to the same velocity as the truck. Let's call this time "t1".
We can use the equation of motion:
v = u + at
Where:
v = final velocity (15 m/s for the truck)
u = initial velocity (0 m/s for the car)
a = acceleration (3.1 m/s^2 for the car)
t = time (t1)
Rearranging the equation and substituting the given values:
15 m/s = 0 m/s + 3.1 m/s^2 * t1
Simplifying, we find:
t1 = 15 m/s / 3.1 m/s^2
Now, we need to find the time it takes for the car to catch up and pass the truck after accelerating. Let's call this time "t2".
Since the car is moving at a constant velocity after accelerating, we can use the equation:
v = d / t
Where:
v = velocity (15 m/s for the car)
d = distance (the distance the truck has traveled at this point)
t = time (t2)
The distance the truck has traveled is equal to its velocity multiplied by the time it has been moving since the car started accelerating, which is t1:
d = 15 m/s * t1
Substituting the values:
t2 = d / v
t2 = (15 m/s * t1) / 15 m/s
Simplifying, we find:
t2 = t1
Therefore, the time it takes for the car to pass the truck is the same as the time it takes for the car to accelerate to the truck's velocity, which we found to be t1.