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.
You divide the distance traveled by the time.
7.5
Yes, many swimmers can even swim 500 meters in 7 minutes.
20 meters per second
1 day = 1440 minutes ⇒ 720 meters per day = 720 meters in 1440 minutes ⇒ 720 ÷ 1440 meters per minute = 1/2 meter per minute = 0.5 meters per minute
0.14
0.14 m/s2 A+
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.
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.
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.
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.
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.
you doing homework???
To find the acceleration, we first need to determine the deceleration since the car is stopping. Using the equation of motion ( v^2 = u^2 + 2as ), where ( v = 0 ) m/s (final velocity), ( u = 19 ) m/s (initial velocity), ( s = 31 ) m (stopping distance), we can find the deceleration. Then, since acceleration is the magnitude of deceleration, the answer would be the calculated deceleration value.
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.
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.
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).