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Each term in the equation has dimensions of velocity-squared (remember "a" here is acceleration which is velocity divided by time, so "as" is velocity x distance / time = velocity squared).
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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!!
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
How do you find velocity in physics?
Velocity can be found either by following formulas, v2 = u2 - 2as or v = distance by time,etc
How to calculate the acceleration of falling body?
TO CALCULATE ACCELERATION OF BODY that is 'g' / 'a' use formula:a=v-u/t where u=0m/s.s=ut+1/2at2.v2-u2=2as or v2-u2=2gh.
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 is nweton first equation of motion?
the first equation of motion is v=u+at the second equation of motion is s=ut+1/2 at2 the third equation of motion is 2as=v2-u2
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 is the LCM of u2-1and u2 plus 9u plus 8?
The LCM is u3 + 8u2 - u - 8u2 - 1 = (u -1)(u + 1)u2 + 9u + 8 = (u + 8)(u + 1)Thus, the LCM = (u - 1)[(u + 1)(u + 8)] = (u - 1)(u2 + 9u + 8).
How do you calculate the initial speed?
v2 = u2+ 2as where v squares is the final velocity , u squared is the initial velocity , a is the acceleration and s is the distance travelled. If it is free fall take a = 10m/s squared ( as gravity ).
A ball is thrown vertically upwards with velocity of 19.6ms what is the maximum height to which it will reach?
apply the formula: v2 - u2 = 2as. Here v= 0; u = 19.6m/s; a = -g ,find s and that's max. heigth
How do you factor completely u2-2u plus 1?
u2 - 2u +1 = u2 - u - u + 1 = u*(u - 1) - 1*(u - 1) = (u - 1)*(u - 1) = (u - 1)2
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!!
