Linear Motion Formulas for an Inclined Plane

[S313]

Homework Statement

The initial speed of a ball up an inclined plane is 3 m/s upwards. The ball reaches its highest point after 1.2 seconds. a) find the acceleration. b) how far has the ball gotten during this time? c) how far has the ball rolled after 0.4 s and after 2.4 seconds?

Homework Equations

The problem is from the first chapter of my physics book, which has only introduced the 4 linear motion formulas (with constant acceleration): v = v0 + at, s = v0t + 1/2at^2, s = ((v0 + v)/2)t and v^2 - v0^2 = 2as.
So here, v0 = 3 m/s and t = 1.2 s

3. The Attempt at a Solution
For part a) I got: a= v/t = 3/1.2 = 2,5 m/s^2
For part b) I used the s = v0t + 1/2at^2 = (3)(1.2) + (1/2)(2.5)(1.2)^2 = 3.6 + 1.8 = 5.4 m
For part c) I couldn't work out the answer.

The final answer is 1.8 m according to the book (I get 5.4), which means that v0 should be 0 (in this way s = (1/2)at^2 will only be used as v0t = 0. But why is v0 = 0? the velocity at the highest point is 0, but initial velocity is supposed to be 3 m/s upwards? why is this ommitted?

 

[BvU]

Hello S,

Are you sure the acceleration and the initial speed have the same sign ?

 

[S313]

Hello S,

Are you sure the acceleration and the initial speed have the same sign ?

Thanks BvU! Happy to be here!
Even if we assume that v0 = -3 m/s, the acceleration will be: a = -v0/t = -3/1.2 = -3.6. Plugging this into s = v0t + 1/2at^2 will yield: -3.6 - 1.8 = -5.4 m. I won't get only 1.8 m. To get the final answer to be 1.8 m, I should assume v0=0 (then v0t = 0) and the only 1/2at^2 becomes 1.8 m, but the problem at hand clearly states initial velocity 3 m/s.

 

[Chandra Prayaga]

BvU's question is very important. It will become clear once you draw the diagram. In the diagram you can assign positive and negative directions to all the quantities. The confusion you are getting into is because there is no diagram to guide your thinking.

 

[BvU]

v = v0 + at

So, if after 1.2 seconds the speed has dropped to zero, the equation reads $$v(1.2 \;{\rm s}) = 0 \; {\rm m/s} = v_ 0 + a \; 1.2\; {\rm s } $$ and with $v_0 = 3 \; {\rm m/s } \$ given, $a$ is easily calculated to be ...

 

[S313]

So, if after 1.2 seconds the speed has dropped to zero, the equation reads $$v(1.2 \;{\rm s}) = 0 \; {\rm m/s} = v_ 0 + a \; 1.2\; {\rm s } $$ and with $v_0 = 3 \; {\rm m/s } \$ given, $a$ is easily calculated to be ...

a = -3/1.2 = -2.5 m/s^2, but how will this lead to finding how far up the slope it went?

 

[BvU]

s = v0t + 1/2at^2

 

[S313]

Yes:
s = v0t + 1/2at^2
= (-3)(1.2) + (1/2)(-2.5)(1.2)^2
= -3.6 - 1.8
= - 5.4 m, but this is not the answer. The answer is 1.8 m according to the book.

 

[BvU]

$v_0 = +\,3 \; {\rm m/s}$

 

[S313]

$v_0 = +\,3 \; {\rm m/s}$

but if v0 = + 3 m/s, then:
a = 3/1.2 = + 2.5 m/s^2
and
s = v0t + 1/2at^2
= 3(1.2) + (1/2)(2.5)1.2^2
= 3.6 + 1.8 = 5.4 m

 

[S313]

Could it be that since the slope is upwards, the acceleration has to have opposite sign of v0? if we choose the + x-axis to be upwards along the slope, then v0 will be positive, while acceleration will slow down and be negative (opposite to v0)?
If so, then s= 3(1.2) + (1/2)(-2.5)(1.2)^2 = 3.6 - 1.8 = 1.8 m

 

[jbriggs444]

Could it be that since the slope is upwards, the acceleration has to have opposite sign of v0?

That is what @BvU was suggesting in post #2.