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A speed boat is 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 information is not consistent.
Let u = initial velocity = 25 m/s
v = final velocity = 0 m/s
s = distance travelled = 100 m t = time - in seconds = 180 s
a = acceleration (so deceleration = -a).
Eqn 1
v = u+at => a = (v - u)/t = (0 - 25)/180 = -25/180 = -0.13888... m/s^2
Eqn 2
s = vt-0.5at^2 => a = (vt - s)/(0.5t^2)
= -100/(0.5*180^2) = -0.006173 (approx) m/s^2.
Eqn 3
s = ut+0.5at^2 => a = (s - ut)/(0.5t^2)
= (100 - 25*180)/(0.5*180^2) = -0.2716 m/s^2
Eqn 4
v^2-u^2 = 2as => a = (v^2 - u^2)/2s
= -25^2/(2*100) = -3.125 m/s^2
Take your pick.
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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
What is the rate of deceleration for a speed boat moving at a velocity of 25 meters per second which runs out of gas and drifts to a stop 3 minutes later 100 meters away?
0.14 m/s2 A+
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?
you doing homework???
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).
