60 Km/h = 60 × 1000 = 60,000 m/s
Acceleration = Change in velocity / Change in time
so , a = v - u / t
.'. a = 60,000 - 0 / 10
.'. a = 60,000 / 10
.'. a = 6,000 / 1
.'. a = 6,000
Then the Acceleration is 6,000 m/s2 ..........
OR :-
10 seconds = 10'÷ 60 = 1/6 hours
Acceleration = Change in velocity / Change in time
so , a = v - u / t
.'. a = 60 - 0 / (1/6)
.'. a = 60 / (1/6)
.'. a = 60 × 6
.'. a = 360
Then the Acceleration is 360 km/h2 ..........
So the Acceleration is either 6,000 m/s2 or 360 Km/h2
The gravitational acceleration would be the change in velocity divide by the time required. For this example 8.15 m/sec divided by 5 seconds yields 1.63 m/s2. (The actual average lunar gravity is estimated at 1.622 m/sec2.)
In two seconds of fall, the speed increases 19.6 meters (64.4 feet) per second. The magnitude of velocity increases by that amount, while the direction of velocity doesn't change.
Acceleration has a dimensionality of length/time^2, so if you were measuring the distance in meters and the time in seconds, the acceleration would be m/s^2.
The acceleration of the roller coaster would be 88 MPH. This was figured out by doing math.
Answer: v=u + at v (Velocity) = u (Starting velocity) + a (acceleration) x t (time) So, starting from stationary (u=0), the velocity is simply a x t e.g. if the acceleration is 5mph per second per second, after 10 seconds you would be travelling at 50mph. Answer: The above is for constant acceleration. In the case of variable acceleration, integration has to be used.
The gravitational acceleration would be the change in velocity divide by the time required. For this example 8.15 m/sec divided by 5 seconds yields 1.63 m/s2. (The actual average lunar gravity is estimated at 1.622 m/sec2.)
Acceleration due to gravity is 9.8m/s/s, which is the same as 9.8m/s2. An acceleration of 9.8m/s/s means that with each passing second, the velocity of the skydiver increases by 9.8m/s. Therefore, after two seconds. a skydiver's velocity would be 19.6m/s. The acceleration will continue at 9.8m/s/s until the skydiver reaches terminal velocity, at which point the weight of the skydiver and the air resistance will be balanced, so the net force acting on the skydiver will be zero, at which point there will be no further acceleration.
Acceleration = Change in speed/Time Time = Change in Speed/Acceleration = 65mph/20mph per s = 65/20 seconds = 3.25 seconds
In two seconds of fall, the speed increases 19.6 meters (64.4 feet) per second. The magnitude of velocity increases by that amount, while the direction of velocity doesn't change.
Acceleration has a dimensionality of length/time^2, so if you were measuring the distance in meters and the time in seconds, the acceleration would be m/s^2.
To answer this question we would need to know the acceleration, which is not provided.
Yes. The acceleration is directly proportional to the objects mass.For objects with constant mass however, the acceleration will remain constant.
Divide the difference in speed by the time it takes. This will give you the average acceleration for that time period.
known to be seconds pendulum,the length would be almost 1m when acceleration due to gravity is 9.8m/s2
55/6 = 9.16666...mph/s
As the speed increases the centripetal acceleration increases. This this acceleration is caused by the normal force, then the normal force increases. Frictional force however remains equal to the weight. As the normal increases the maximum frictional force would increase. However, the amount of friction needed would still only be the rider's weight. In other words the frictional force would be less than the maximum available frictional force.
Since the derivative of velocity is acceleration, the answer would be technically 'no'. Here is why: v = 0 v' = 0 = a Or in variable form... v(x) = x v(0) = 0 v'(0) = 0 = a You can "trick" the derivative into saying that v'(x) = 1 = a (since the derivative of x = 1) and then stating v'(0) = 1 = a... but that is not entirely correct. Acceleration is a change over time and is measured at more then one point (i.e. the acceleration of this body of matter is y from time 1 to 5) unless using derivatives to form the equation of the acceleration line/curve. If an object has a constant acceleration of 1, then the velocity is constantly increasing over that time. Using the equation discussed above and looking at acceleration over time, at 0 seconds, acceleration is 0 and so is velocity, but from 0-1 seconds acceleration is 1 and velocity is 1 as well. 0-2 seconds, acceleration is 1, but velocity would be 2 (at the end of 2 seconds).