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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 your null hypothesis is Ho u1 equals u2?
yes.
After stopping at a station a train accelerates at a steady rate of 0.25ms-2 At what speed is the train moving after it has traveled 1.0 km?
Using v2 - u2 = 2as where, v = final velocity u = initial velocity a = acceleration s = distance travelled we get v2 = u2 + 2as Substituting the values, we get v2 = 02 + (2 x 0.25 x 1000) so v2 = 500, therefore v = 22.36 The train will be travelling at 22.36 ms-1.
What speed must an object leave the ground to travel 2.5 meters up?
Using the physics formula: v2 = u2 + 2as v - final velocity u - initial velocity a - acceleration s - distance When the object leaves the ground it is being slowed down by gravity until it stops so the acceleration equals -9.8m/s2 and as it will stop at 2.5 meters v = 0. U is the value we want to find out: 02 = u2 + (2 *-9.8*2.5) 49 = u2 So the object must leave the ground at 7m/s. By the way this is ignoring air resistance.
What factors are involved in determining stopping distance of a toy car?
arent u a science student? its simple. time,initial velocity,accelaration the formula is; v2=u2+2as here v is final velocity u is innitial velocity a is accelaration s will be3 the stopping distance okkkk!!
