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A speed boat is moving at a velocity of 25 ms runs out of gas and drifts to a stop 3 minutes later 100 meters away What is the rate of deceleration?
0.14
How A speed boat moving at a velocity of 25 ms runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its rate of deceleration?
To find the rate of deceleration, we first need to convert the time from minutes to seconds. 3 minutes = 180 seconds. Next, we use the equation of motion: final velocity^2 = initial velocity^2 + 2 * acceleration * distance. Since the boat comes to a stop, the final velocity is 0 m/s. Substituting the values and solving for acceleration, we get a deceleration rate of 0.139 m/s^2.
A speed boat moving at a velocity of 25 m/s runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its rate of deceleration?
The initial velocity is 25 m/s, final velocity is 0 m/s, and the distance covered is 100 meters. Convert 3 minutes to seconds (3 minutes = 180 seconds). Use the equation v^2 = u^2 + 2as, where v is final velocity, u is initial velocity, a is acceleration, and s is distance. Solve for acceleration (deceleration in this case) to find it to be -0.3472 m/s^2. The negative sign indicates deceleration.
A speed boat moving at a velocity of 25 ms runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its rate of deceleration?
The initial velocity was 25 m/s and it came to a stop over a distance of 100 meters in 3 minutes (180 seconds). To find the deceleration, use the equation v^2 = u^2 + 2as, where v is final velocity (0 m/s), u is initial velocity (25 m/s), a is acceleration, and s is distance. Solving for acceleration gives you approximately -0.35 m/s^2, indicating the boat's rate of deceleration.
When A speed boat moving at a velocity of 25 ms runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its rate of deceleration?
The boat decelerated from 25 m/s to 0 m/s in 3 minutes (180 seconds). Using the equation of motion (a = \frac{(v_f - v_i)}{t}), where (a) is acceleration, (v_f) is final velocity, (v_i) is initial velocity, and (t) is time, the acceleration (deceleration in this case) is (\approx -\frac{25}{180} \approx -0.14) m/s².
In A speed boat moving at a velocity of 25 ms runs out of gas and drifts to a stop 3 minutes later 100 meters away. What is its rate of deceleration?
To find the rate of deceleration, we need to first convert the time to seconds. 3 minutes is 180 seconds. Then, we can use the formula for acceleration: acceleration = change in velocity / time taken. The change in velocity is from 25 m/s to 0 m/s, which is -25 m/s (negative because it's decelerating). So, the rate of deceleration is -25 m/s / 180 s = -0.139 m/s².
What is the deceleration of a car traveling with a velocity of 30 meters per second slowing down to a velocity of 10 meters per second within 10 seconds?
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If a car moving at an initial velocity of 19 meters per sec has a stopping distance of 31 meters what is the magnitude of its acceleration?
Average speed during the deceleration is 1/2(19 + 0) = 9.5 meters per second.Time of deceleration is (31 / 9.5) seconds.Magnitude of deceleration is (change of speed) / (deceleration time) = 19 / (31/9.5) = (19 x 9.5) / 31 = 5.823 m/s2(The acceleration is the negative of this number.)
What are the units to a deceleration problem?
The units for deceleration are typically meters per second squared (m/s^2) in the metric system or feet per second squared (ft/s^2) in the imperial system. Deceleration represents a decrease in velocity over time.
What deceleration is needed to bring a car with a velocity of ms-1 to rest in 80 meters?
The deceleration needed to bring a car to rest can be calculated using the equation: ( v^2 = u^2 + 2as), where (v) is the final velocity (0 m/s), (u) is the initial velocity (given as unknown), (a) is the deceleration, and (s) is the displacement (80 meters). Solving for (a), the deceleration needed would be about 2.52 m/s^2.
What is deceleration measured in?
Deceleration means to decrease the velocity. The SI unit is the same as acceleration. In SI units, acceleration is measured in meters/second² (m·s-2).
When throwing a ball straight up when in the air with an initial velocity of 10 meters per second what high will it go and how long will it take to return to the ground?
The initial velocity is 10 meters/sec and is thrown up against the gravitational pull of the earth. This means that the ball is experiencing a deceleration at the rate of 9.8 meters/sec/sec to bring its final velocity to zero. v^2 - u^2 = 2gs where u is the initial velocity, v the final velocity, g is the acceleration or deceleration, and s is the distance traveled. 0^2 - 10^2 = 2 x (-9.8) x s -100 = -19.6s 100 = 19.6s s = 100/19.6 = 5.102 meters Now v = u + gt where v is the final velocity, u is the initial velocty, g is the acceleration or deceleration, and t is the time. When the ball is thrown up with 10 meters/sec velocity it is acted upon by the deceleration of gravity until its velocity becomes zero. So 0 = 10 - 9.8t or 9.8t = 10 t = 1.020 seconds The time for the ball to go up is 1.020 seconds and the same time is taken for the ball to come back for a total of 2.040 seconds.
