question 1 ..A 5 kg rock is thrown vertically upward at 12 m/s. How high did the rock rise?
question 2 The driver of a truck traveling at 126 km/h on a highway sees a sign which indicated that 800 m after the sign the maximum speed will be 90 km/h . When the truck passes the sign the driver steps on the brake until the truck slows to the new speed limit in the required distance. Assuming the acceleration is constant, how long does it take for the truck to decelerate?
1. Vf^2 = Vo^2 + 2g*h.
Vf = 0.
Vo = 12 m/s.
g = -9.8 m/s^2.
h =
2. V = 90,000m/3600s = 25 m/s.
Vo = 126,000m/3600s = 35 m/s.
V = Vo + a*t.
a = (V-Vo)/t = (25-35)/t = -10/t.
V^2 = Vo^2 + 2a*d.
25^2 = 35^2 + 2(-10/t)*800.
625 = 1225 + (-16,000)/t.
-600 = -16000/t.
600t = 16000.
t = 26.7 s.
For question 1, to find the height the rock rose, we can use the laws of motion. We know the initial upward velocity (u) of the rock is 12 m/s, and the acceleration due to gravity (g) is approximately 9.8 m/s². The final upward velocity (v) of the rock when it reaches its highest point is 0 m/s. We can use the following equation:
v² = u² + 2as
Where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Rearranging the equation to solve for s:
s = (v² - u²) / (2a)
Plugging in the values we know:
s = (0 - (12)²) / (2(-9.8))
s = -144 / (-19.6) = 7.35 meters
Therefore, the rock rose to a height of approximately 7.35 meters.
For question 2, to find the time it takes for the truck to decelerate, we need to use the equations of motion. We know the initial speed (u) of the truck is 126 km/h, the final speed (v) is 90 km/h, and the distance (s) over which it decelerates is 800 m.
First, we need to convert the speeds from km/h to m/s. Since 1 km/h is equal to 0.2778 m/s:
u = 126 km/h * 0.2778 m/s = 35 m/s
v = 90 km/h * 0.2778 m/s = 25 m/s
Now, we can use the equation:
v = u + at
Since the acceleration (a) is negative (due to deceleration), we can rearrange the equation:
t = (v - u) / a
To find the deceleration (a), we can use the equation:
a = (v² - u²) / (2s)
Plugging in the values:
a = (25² - 35²) / (2 * 800)
a = (625 - 1225) / 1600
a = -600 / 1600 = -0.375 m/s²
Now, we plug this value of acceleration into the time equation:
t = (25 - 35) / -0.375
t = -10 / -0.375 = 26.67 seconds
Therefore, it takes approximately 26.67 seconds for the truck to decelerate.