v = u + at
a = (v - u) / t
u = 0 , v = 18 ms-1 , t = 30 s
a = 18 / 30 = 0.6 ms-2
0.712 m/s^2
Speed = (acceleration) x (time) = 9 x 8 = 72 ft/sec
A force. This leads to the equation; force = mass x acceleration Unbalanced(:
Acceleration is the change in velocity divided by the time it takes to make that change;a= delta v/delta tAs you can see, the change in velocity for both vehicles is the same (10km/h), and the question states that they do it in the same amount of time. So the acceleration is the same;delta v1/delta t1 = delta v2/delta t2 = a.The fact that the motorcycle is already at a much higher speed is irrelevant.
a few things missing: - How fast is the acceleration? - How long does it take to go from 4ms to 20ms?
the acceleration of the car is 1.179m/s/s given the formula provided.
0.712 m/s^2
60.912 meters in that time
Speed = (initial speed) plus (acceleration) x (time) = 0 + (5) x (3) = 15 meters per second.
Acceleration = (change in speed) / (time for the change)3 = (30) / (time)3 x (time) = 30time = 30/3 = 10 seconds
Speed = (acceleration) x (time) = 9 x 8 = 72 ft/sec
Since , V = u + at, we get , a = v - u /t = 402.3 - 0 /9.013 = 44.6355264617 ms-2 Therefore, acceleration = 44.6355264617 ms-2
A force. This leads to the equation; force = mass x acceleration Unbalanced(:
So it's acceleration is 4m/s2. So at any point because it says uniformly, it will be accelerating at 4m/s2 each second
Acceleration is the change in velocity divided by the time it takes to make that change;a= delta v/delta tAs you can see, the change in velocity for both vehicles is the same (10km/h), and the question states that they do it in the same amount of time. So the acceleration is the same;delta v1/delta t1 = delta v2/delta t2 = a.The fact that the motorcycle is already at a much higher speed is irrelevant.
No a boday while rest cannot be in acceleration because of the momentum of the body ...........................
Acceleration cannot be measured in metres per second. There is, therefore a fundamental problem with the question.