Wiki User
∙ 13y agoAcceleration is the time rate of change of velocity. If velocity is constant, then
acceleration is zero.
Note:
"100 km per h for 10 seconds" is a constant speed, but not necessarily a constant
velocity, since we're told nothing about the direction. If the car moves in a perfectly
straight line during those 10 seconds, then its velocity is constant. If it makes a curve,
then its velocity is not constant even though its speed is, and there is acceleration.
Wiki User
∙ 13y agoThe answer is very simple. The words "constant velocity" are the definition of zero acceleration.
Constant velocity is a measure of distance traveled per unit of time at a uniform speed, such as miles per hour or feet per second. Constant acceleration is a measure of a continuing increase in velocity per unit of time, as when a car speeds up from 30 miles per hour to 40 miles per hour in 5 seconds, then from 40 miles per hour to 50 miles per hour during the next 5 seconds. It will then have had a constant acceleration of 10 miles per hour per 5 seconds.
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.
Assuming that acceleration is constant during that time, just divide the change in speed by the time.
what is the change in speed or velocity? average acceleration will be change in speed or velocity divided by time taken (4 seconds in ur case)
The answer is very simple. The words "constant velocity" are the definition of zero acceleration.
Yes, velocity is acceleration x time. If acceleration is the same, velocity can be different as it changes with time. For example a car accelerating with constant acceleration will have a different velocity after 5 seconds than it will have at 2 seconds.
No, a skydiver's acceleration remains constant as they fall towards their terminal velocity. This is because terminal velocity is the point at which the forces of gravity and air resistance are balanced, resulting in a constant 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).
Answer This occurs when an object istraveling in one direction but has an acceleration in the opposite direction, which means it is decreasing in speed. For a given period of time, the speed has decreased. Acceleration is the change of velocity per second. T1 = 5 seconds V1 = 100mph T2 = 10 seconds V2= 50 mph Acceleration = (V2 - V1 ) / (T2 - T1) = (50-100)/(10-5) = -10 ft/sec/sec Positive(+) acceleration means an object will be going faster over an interval of time.
To calculate acceleration, you need to know the initial velocity of the car and its final velocity after 6.8 seconds. The acceleration can be found using the formula: acceleration = (final velocity - initial velocity) / time.
Constant velocity is a measure of distance traveled per unit of time at a uniform speed, such as miles per hour or feet per second. Constant acceleration is a measure of a continuing increase in velocity per unit of time, as when a car speeds up from 30 miles per hour to 40 miles per hour in 5 seconds, then from 40 miles per hour to 50 miles per hour during the next 5 seconds. It will then have had a constant acceleration of 10 miles per hour per 5 seconds.
I'm assuming you are not talking about constant acceleration, which in itself iscontinuous. If you are talking about constant (meaning it's just a number, and not a function itself) acceleration, than just use the formula:v = i + at, wherev = velocity, i = initial velocity, a = acceleration, and t = timeOtherwise, you need to have some calculus knowledge. A velocity function is just the antiderivative (integral) of an acceleration function.Say, acceleration was given as:a = 2ttake the integral of thatv =∫ 2t dt = (2/2)t2+c =t2+cIn this case, the c is the initial velocity.Example:An object moves with an acceleration determined by a=t+3 (in m/s^2). Find the velocity of the object after 10 seconds have passed, given that the object has an initial velocity of 2 m/s.a = t + 3v =∫( t + 3) dt = (1/2)t2 + 3t + c, with c being initial velocityv(10 seconds) = (1/2)(10)2 + 3(10) + (2) = 82 m/s
To calculate acceleration between 6 and 9 seconds, you need to find the change in velocity during that time interval and then divide it by the time taken. The formula for acceleration is acceleration = (final velocity - initial velocity) / time. Plug in the velocities at 6 seconds and 9 seconds into the formula to get the acceleration.
Acceleration is the rate of change in velocity (speed) Thus if your speed is constant (50 miles per hour) your acceleration is zero
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.
Assuming that acceleration is constant during that time, just divide the change in speed by the time.