Which Distance Formula?

Question

A force of 17N is applied to a 6KG block which is at rest. Find:
A] Acceleration
B] Velocity
C] Distance After 9 seconds.

I solved A with the formula Force divided by Mass. [a=F/m]
17/6=2.83 m/s2

Solved B with the formula: Initial Velocity plus Acceleration times Time [Vf=Vi+a(t)]
0 + 2.83(9) = 25.47m/s

Now, for the distance (c) i have two formulas. the first one is:
Distance = Velocity * Time
and
Distance = Initial Velocity (time) + 1/2 a (time)2


Which formula is the right one to use?

Answer 1

Distance = Velocity * Time
will give you a result of 229.23m

Distance = Initial Velocity (time) + 1/2 Acceleration (Time)^2
will give you a result of 114.615m

The first formula assumes constant velocity with no acceleration. Since you have already calculated that the acceleration is not equal to zero, then you don't have constant velocity. Therefore use the second formula.

Answer 2

Physics Distance Formula

Answer 3

Distance Formula Physics

Answer 4

Since the velocity is changing from zero to a velocity of 25.47 m/s ,

you cannot use the formula " Distance = velocity x time" as it is.

However, you can change it considering the change in the velocity as

"DISTANCE = AVERAGE VELOITY x TIME

Then the average velocity is [25.47 + 0] / 2 = 12.735 m/s

Distance = 12 .735 x 9 = 114.615m.
=================================
You can use the equation
Distance = Initial Velocity (time) + 1/2 a (time) ^2
Distance = 0 + [1/2] * 2.83 * 9^2 = 114.615 m
=============================

Answer 5

You're doing a good job so far!

When you learn an equation, you should also learn the assumptions that went into making it. Distance = velocity * time assumes that the velocity is constant, which it isn't here. Only use that equation when there is no acceleration.

x = v0 t + 1/2 a t^2 assumes constant acceleration. The acceleration in this problem is constant (usually it will be in first year physics) so you can use it. You know the acceleration is constant because the force is constant.

Answer 6

For the best answers, search on this site https://shorturl.im/RNmTn

Hi Samantha, The distance formula is d=sqrt[(x2-x1)^2+(y2-y1)^2]. We need the distance from a point with an x-coordinate of 2 to the point (-2,-1) to be 5, so we let (x1,y1)=(-2,-1), (x2,y2)=(2,y), and d=5, and then solve for y: sqrt[(2 - -2)^2 + (y - -1)^2] = 5 sqrt[4^2 + (y+1)^2] = 5 (y+1)^2 + 4^2 = 25 y^2 + 2y + 1 + 16 = 25 y^2 + 2y - 8 = 0 (y-2)(y+4) = 0 y = 2, y = -4 So, the points that satisfy the given requirements are (2,2) and (2,-4). I hope that helps :)

Answer 7

Distance = Velocity * Time is for uniform velocity. Use your second formula for acceleration.

Answer 8

the 2nd formula can only be used under circumstances of uniform acceleration. Since acceleration is not specified to be uniform in this case, i'ld use the first one i.e. d=v*t

N.B. not sure, u'ld hv to check that out.