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What else can I help you with?
Can a body have constant acceleration and zero velocity?
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).
When acceleration is 0 why does velocity have to be at a maximum?
It doesn't. If acceleration is zero, that just means that velocity isn'tchanging ... the motion is in a straight line at a constant speed.
What acceleration does a rocket need to reach a speed of 230meters per second at a height of 1.0km?
To calculate the acceleration of the rocket, you can use the kinematic equation: (v^2 = u^2 + 2as), where (v) is the final velocity (230 m/s), (u) is the initial velocity (0 m/s assuming the rocket starts from rest), (a) is the acceleration, and (s) is the displacement (1000 m). Rearranging the equation to solve for acceleration, you'd get (a = (v^2 - u^2) / 2s). Plugging in the values, you'd get (a = (230^2 - 0^2) / (2 * 1000) = 264.5 m/s^2). The rocket would need an acceleration of 264.5 m/s^2 to reach a speed of 230 m/s at a height of 1.0 km.
When velocity is zero and acceleration is less than zero does the speed increase?
if acceleration is <0 and velocity =0 then you got the handbrake on
Can your speed be 0 while your acceleration non zero?
Yes, it is possible for the speed to be 0 while the acceleration is non-zero. This occurs when the object is momentarily at rest (speed is zero) while still experiencing acceleration due to a change in its velocity.
When an object is accelerating constantly how to find the acceleration?
When the acceleration is constant then s= 1/2 at^2 and the acceleration is a=2s/t^2.
How do you calculate acceleration when given velocity and distance?
v2 - u2 = 2as so that a = (v2 - u2)/2s where u = initial velocity v = final velocity s = distance a = acceleration
If the universe was expanding with 0 acceleration what would the age of the universe be?
The universe simply cant expand at 0 acceleration.
What is the perpendicular bisector equation of the line joined by the points of s 2s and 3s 8s?
Points: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular equation: y-5s = -1/3(x-2s) => 3y-15s = -x+2s => 3y = -x+17s Perpendicular bisector equation in its general form: x+3y-17s = 0
What is the quantum set for 2s orbital of aluminum?
For 2s orbital:Principal Q.N. (n) = 2Azimuthal Q.N. (l) = 0Magnetic Q.N. (ml) = 0Spin Q.N. (ms) = -1/2 and +1/22, 0, 0, +/- 1/2
A sprinter accelerates from 1 ms to 3ms in 2s?
And what is the question? If you want the average acceleration for that time, divide the change in velocity, by the time elapsed.
Can a body have constant acceleration and zero velocity?
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).
What is the perpendicular bisector equation joining the points of s 2s and 3s 8s on the Cartesian plane showing work?
Points: (s, 2s) and (3s, 8s) Slope: (8s-2s)/(3s-s) = 6s/2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) => 3y-15s = -1(x-2s) => 3y = -x+17x Perpendicular bisector equation in its general form: x+3y-17s = 0
When acceleration is 0 why does velocity have to be at a maximum?
It doesn't. If acceleration is zero, that just means that velocity isn'tchanging ... the motion is in a straight line at a constant speed.
What is the perpendicular bisector equation of the line whose endpoints are s 2s and 3s 8s?
Endpoints: (s, 2s) and (3s, 8s) Midpoint: (2s, 5s) Slope: 3 Perpendicular slope: -1/3 Perpendicular bisector equation: 3y = -x+17s or as x+3y-17s = 0
What is the perpendicular bisector equation of a line joined by the points of s 2s and 3s 8s showing key stages of work?
It is found as follows:- Points: (s, 2s) and (3s, 8s) Slope: (2s-8s)/(s-3s) = -6s/-2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) Multiply all terms by 3: 3y-15s = -1(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0
How do you find acceleration falling object with only speed and distance?
a = (v2 - u2)/2s where a is the acceleration between the initial point in time and the final point in time, u is the initial velocity v is the final velocity s is the distance travelled
