A robot is seen to accelerate 125855 km/hr2 over a distance of 62.4 m. If it takes the robot 16 s, what is the final velocity in m/s? (2 decimal places)
To find the final velocity, we can use the equation:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity (which we assume to be 0 m/s as the robot starts from rest), a is the acceleration, and s is the distance.
Given that the acceleration is 125855 km/hr², we need to convert it to m/s²:
125855 km/hr² = 125855 * (1000 m/1 km) * (1 hr/3600 s)²
= (125855 * 1000 * 1) / (3600 * 3600) m/s²
= 34.96 m/s² (rounded to two decimal places)
Plugging the values into the equation, we have:
v^2 = 0^2 + 2 * 34.96 m/s² * 62.4 m
= 2 * 34.96 m/s² * 62.4 m
= 4378.69 m²/s² (rounded to two decimal places)
Taking the square root of both sides of the equation, we find the final velocity v:
v = sqrt(4378.69 m²/s²)
= 66.18 m/s (rounded to two decimal places)
Therefore, the final velocity of the robot is approximately 66.18 m/s.
A ball is thrown up into the air with a speed of 12.7 m/s from a 11.9 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 can use the equation:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity, a is the acceleration (which is the acceleration due to gravity, -9.8 m/s²), and s is the vertical distance traveled by the ball.
Given that the ball is thrown up into the air with a speed of 12.7 m/s from an initial height of 11.9 m, we can see that the distance traveled by the ball is the sum of the initial height and the final height (which is 0 m since the ground is at 0 height).
Using the equation s = u*t + 0.5*a*t^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time, we can find the time it takes for the ball to hit the ground.
0 = 11.9 m + 0.5 * (-9.8 m/s²) * t^2,
which simplifies to:
4.9 * t^2 = 11.9 m.
Solving for t, we have:
t^2 = 11.9 m / 4.9 m/s²
t^2 = 2.429 m/s²,
t ≈ √2.429 s
t ≈ 1.56 s (rounded to two decimal places).
Now that we know the time it takes for the ball to fall, we can plug it into the equation v = u + at:
v = 12.7 m/s + (-9.8 m/s²) * 1.56 s
v ≈ 12.7 m/s + (-15.288 m/s)
v ≈ -2.59 m/s (rounded to two decimal places).
Since the velocity is negative, it indicates that the ball is moving downwards. Therefore, the ball hits the ground with a velocity of approximately -2.59 m/s.
A gorilla is running 1.8 m/s when it starts to accelerate at 0.70 m/s2 . How long will it take the gorilla to make it 19.9 m? (answer to 2 decimal places)
To find the time it takes for the gorilla to travel a distance of 19.9 m, we can use the equation:
s = ut + 0.5at^2,
where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Given that the initial velocity of the gorilla is 1.8 m/s and the acceleration is 0.70 m/s², and the distance to be traveled is 19.9 m, we can substitute these values into the equation:
19.9 m = 1.8 m/s * t + 0.5 * 0.70 m/s² * t^2,
Rearranging the equation, we get:
0.35 t^2 + 1.8 t - 19.9 = 0.
This is a quadratic equation, which can be solved using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a,
where a = 0.35, b = 1.8, and c = -19.9.
Plugging in the values, we have:
t = (-1.8 ± √(1.8^2 - 4 * 0.35 * -19.9)) / 2 * 0.35
= (-1.8 ± √(3.24 + 27.58)) / 0.70
= (-1.8 ± √30.82) / 0.70.
Since we are looking for a positive time, we can ignore the negative value in ±. Therefore, we have:
t = (-1.8 + √30.82) / 0.70
≈ 3.17 s (rounded to two decimal places).
Thus, it will take the gorilla approximately 3.17 seconds to travel a distance of 19.9 m.
A spaceship is moving 10.4 m/s. It then accelerates for 5.4s over a distance of 624m. What would the final velocity be? (answers to 2 decimal places)
To find the final velocity of the spaceship, we can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given that the spaceship is initially moving at a velocity of 10.4 m/s, and it accelerates for 5.4 s with a distance of 624 m, we can calculate the acceleration.
The equation relating distance, initial velocity, final velocity, and acceleration is:
s = ut + 0.5at^2,
where s is the distance. By substituting the values, we can find the acceleration.
624 m = 10.4 m/s * 5.4 s + 0.5 * a * (5.4 s)^2.
Rearranging the equation, we have:
156 = 56.16 + 14.58a.
Subtracting 56.16 from both sides, we get:
99.84 = 14.58a.
Dividing both sides by 14.58, we find:
a ≈ 6.842 m/s² (rounded to two decimal places).
Now that we know the acceleration, we can calculate the final velocity using the equation:
v = u + at.
v = 10.4 m/s + 6.842 m/s² * 5.4 s
v ≈ 10.4 m/s + 36.9868 m/s
v ≈ 47.39 m/s (rounded to two decimal places).
Therefore, the final velocity of the spaceship would be approximately 47.39 m/s.
A car is seen driving 45 m/s [Forward]. After 49 s, it is seen going 19 km/h [backward]. What is the car's acceleration in m/s2 ? (put your answer to 2 decimal places)
To find the car's acceleration, we can use the equation:
a = (v - u) / t,
where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time.
Given that the car is initially moving forward at a velocity of 45 m/s and after 49 s it is moving backward at a velocity of 19 km/h, which we need to convert to m/s, we can calculate the acceleration.
Converting 19 km/h to m/s:
19 km/h = 19 * (1000 m/1 km) * (1 hr/3600 s)
= (19 * 1000 * 1) / 3600 m/s
≈ 5.28 m/s.
Substituting the values into the equation, we have:
a = (5.28 m/s - 45 m/s) / 49 s
= (-39.72 m/s) / 49 s
≈ -0.81 m/s² (rounded to two decimal places).
The negative sign indicates that the car is accelerating in the opposite direction of its initial velocity, which means it is decelerating or slowing down.
Therefore, the car's acceleration is approximately -0.81 m/s².
To find the final velocity, we need to use the formula:
final velocity (v) = initial velocity (u) + acceleration (a) * time (t)
However, in this problem, the initial velocity is not given. Let's assume that the initial velocity is zero since it is not given in the question.
Given:
Acceleration (a) = 125855 km/hr²
Distance (d) = 62.4 m
Time (t) = 16 s
Initial velocity (u) = 0
To convert the acceleration from km/hr² to m/s², we need to use the conversion factor:
1 km/hr² = 0.02778 m/s²
So, now we have:
Acceleration (a) = 125855 km/hr² * 0.02778 m/s²/km/hr² ≈ 3490.16 m/s²
Using the formula, we can calculate the final velocity (v):
v = u + a * t
v = 0 + 3490.16 m/s² * 16 s
v ≈ 55842.56 m/s
Rounding to 2 decimal places, the final velocity is approximately 55842.56 m/s.