A robot is seen to accelerate 125465 km/hr2 over a distance of 85.0 m. If it takes the robot 20 s, what is the final velocity in m/s? (2 decimal places)
To find the final velocity, we can use the formula for acceleration:
a = (v_f - v_i) / t
Where:
a is the acceleration,
v_f is the final velocity,
v_i is the initial velocity (which we assume to be 0 km/hr since it starts from rest),
and t is the time.
Converting the given acceleration from km/hr^2 to m/s^2:
125465 km/hr^2 * (1000 m/1 km) * (1 hr/3600 s) = 34.85 m/s^2
Rearranging the formula to solve for v_f:
v_f = a * t + v_i
Substituting the given values:
v_f = 34.85 m/s^2 * 20 s + 0 m/s
v_f = 697 m/s
So, the final velocity of the robot is 697 m/s.
A car is seen driving 13 m/s [Forward]. After 62 s, it is seen going 39 km/h [backward]. What is the car's acceleration in m/s2 ? (put your answer to 2 decimal places)
To find the acceleration of the car, we need to find the change in velocity and divide it by the time elapsed.
First, let's convert the final velocity from km/h to m/s:
39 km/h * (1000 m / 1 km) * (1 hr / 3600 s) = 10.83 m/s [Backward]
The initial velocity is given as 13 m/s [Forward].
Therefore, the change in velocity is:
Δv = final velocity - initial velocity
Δv = 10.83 m/s - 13 m/s
Δv = -2.17 m/s
Now, we can calculate the acceleration:
a = Δv / t
a = -2.17 m/s / 62 s
a ≈ -0.035 m/s²
Therefore, the car's acceleration is approximately -0.04 m/s².
A ball is thrown up into the air with a speed of 18.3 m/s from a 10.6 m high balcony. With what velocity does it hit the ground? (2 decimal places)
To find the velocity with which the ball hits the ground, we need to consider the motion of the ball as it goes up and comes back down.
First, we can find the time taken for the ball to reach its maximum height. We can use the equation:
v = u + at
Where:
v is the final velocity (0 m/s at the maximum height),
u is the initial velocity (18.3 m/s),
a is the acceleration due to gravity (-9.8 m/s²),
and t is the time taken to reach the maximum height.
Rearranging the equation to solve for t:
t = (v - u) / a
t = (0 m/s - 18.3 m/s) / (-9.8 m/s²)
t ≈ 1.86 s
Next, we can find the distance covered by the ball in the upward journey using the equation:
s = ut + (1/2)at²
Where:
s is the distance covered,
u is the initial velocity,
t is the time,
and a is the acceleration due to gravity.
Substituting the given values:
s = 18.3 m/s * 1.86 s + (1/2) * (-9.8 m/s²) * (1.86 s)²
s ≈ 17.03 m
Now, we can find the remaining distance to the ground by subtracting the distance covered in the upward journey from the initial height:
remaining distance = initial height - distance covered in upward journey
remaining distance = 10.6 m - 17.03 m
remaining distance ≈ -6.43 m
Since the remaining distance is negative, indicating that the ball is below the initial height, we can ignore the negative sign and consider the magnitude of the remaining distance.
Finally, we can calculate the time taken to reach the ground using the equation:
s = ut + (1/2)at²
Where:
s is the remaining distance,
u is the initial velocity (0 m/s when the ball hits the ground),
t is the time taken to reach the ground (which we need to find),
and a is the acceleration due to gravity (-9.8 m/s²).
Rearranging the equation:
t = √((2s) / a)
t = √((2 * 6.43 m) / (-9.8 m/s²))
t ≈ 0.90 s
Finally, we can find the velocity with which the ball hits the ground using the equation:
v = u + at
Where:
v is the final velocity,
u is the initial velocity (0 m/s),
a is the acceleration due to gravity (-9.8 m/s²),
and t is the time taken to reach the ground.
v = 0 m/s + (-9.8 m/s²) * 0.90 s
v ≈ -8.82 m/s
Therefore, the ball hits the ground with a velocity of approximately -8.82 m/s.
A spaceship is moving 5.8 m/s. It then accelerates for 2.9s over a distance of 604m. What would the final velocity be? (answers to 2 decimal places)
To find the final velocity of the spaceship, we can use the formula for velocity:
v = u + at
Where:
v is the final velocity,
u is the initial velocity,
a is the acceleration,
and t is the time.
Given:
Initial velocity (u) = 5.8 m/s
Time (t) = 2.9 s
To find the acceleration, we can use the formula:
s = ut + (1/2)at²
Where:
s is the distance covered,
u is the initial velocity,
t is the time,
and a is the acceleration.
Given:
Distance covered (s) = 604 m
Rearranging the formula to solve for acceleration:
a = (2s - 2ut) / t²
Substituting the given values:
a = (2 * 604 m - 2 * 5.8 m/s * 2.9 s) / (2.9 s)²
a ≈ 39.04 m/s²
Now, we can find the final velocity:
v = u + at
v = 5.8 m/s + 39.04 m/s² * 2.9 s
v ≈ 114.98 m/s
Therefore, the final velocity of the spaceship would be approximately 114.98 m/s.
A gorilla is running 4.3 m/s when it starts to accelerate at 2.20 m/s2 . How long will it take the gorilla to make it 18.4 m? (answer to 2 decimal places)
To find the time it will take for the gorilla to make it 18.4 m, we can use the equation:
s = ut + (1/2)at²
Where:
s is the distance covered,
u is the initial velocity,
t is the time,
and a is the acceleration.
Given:
Initial velocity (u) = 4.3 m/s
Acceleration (a) = 2.20 m/s²
Distance covered (s) = 18.4 m
Rearranging the equation to solve for time (t):
t² + 2u/t - 2s/a = 0
Substituting the given values:
t² + 2 * 4.3 m/s * t - 2 * 18.4 m / 2.20 m/s² = 0
This is a quadratic equation in terms of (t). We can solve it using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Where:
a = 1
b = 2u = 2 * 4.3 m/s = 8.6 m/s
c = -2s/a = -2 * 18.4 m / 2.2 m/s² = -16.727 m
Substituting the values in the quadratic formula:
t = (-8.6 m/s ± √((8.6 m/s)² - 4 * 1 * (-16.727 m))) / (2 * 1)
Calculating the values inside the square root:
t = (-8.6 m/s ± √(73.96 m² + 134.81 m²)) / 2
t = (-8.6 m/s ± √(208.77 m²)) / 2
t = (-8.6 m/s ± 14.46 m) / 2
Now, we can find two possible values for time (t):
t₁ = (-8.6 m/s + 14.46 m) / 2
t₁ = 5.28 s
t₂ = (-8.6 m/s - 14.46 m) / 2
t₂ = -11.53 s
Since time cannot be negative in this context, the gorilla will take approximately 5.28 seconds to cover a distance of 18.4 m.
To find the final velocity of the robot, we can use the equation of motion:
Vf = Vi + a * t
Where:
Vf is the final velocity,
Vi is the initial velocity (assumed to be 0 since it is not given),
a is the acceleration, and
t is the time.
Given:
Acceleration (a) = 125465 km/hr^2
Distance (d) = 85.0 m
Time (t) = 20 s
First, we need to convert the given acceleration from km/hr^2 to m/s^2.
1 km = 1000 m
1 hr = 3600 s
To convert km/hr^2 to m/s^2, we need to perform two conversions:
1. Conversion of kilometers to meters: Multiply by 1000.
2. Conversion of hours to seconds: Multiply by (3600)^2.
125465 km/hr^2 * 1000 m/km * (1 hr/3600 s) * (1 hr/3600 s)
Simplifying the units, we get:
a = 34.85 m/s^2
Now we can substitute the values into the equation to find the final velocity:
Vf = Vi + a * t
Since the initial velocity Vi is assumed to be 0, we can simplify the equation to:
Vf = a * t
Vf = 34.85 m/s^2 * 20 s
Vf = 697.0 m/s
Therefore, the final velocity of the robot is 697.0 m/s.