Linear Motion Formulas for an Inclined Plane

• S313
In summary: However, 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)?Yes, that is correct.
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?

Hello S,

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

BvU said:
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.

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.

gneill, PeroK and Merlin3189
S313 said:
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 ...

BvU said:
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
S313 said:
s = v0t + 1/2at^2

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.

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

BvU said:
##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

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

BvU
S313 said:
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.

BvU

What is linear motion?

Linear motion is the movement of an object along a straight line, with a constant velocity or acceleration.

What is the formula for linear motion?

The formula for linear motion is x = x0 + v0t + 1/2at^2, where x is the final position, x0 is the initial position, v0 is the initial velocity, t is the time, and a is the acceleration.

How do you calculate velocity in linear motion?

The formula for velocity in linear motion is v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is the acceleration, and t is the time.

What is the difference between velocity and acceleration in linear motion?

Velocity is the rate of change of an object's position, while acceleration is the rate of change of an object's velocity. In other words, velocity tells us how fast an object is moving, while acceleration tells us how much an object's velocity is changing over time.

How is linear motion used in real life?

Linear motion is used in many real-life applications, such as a car moving in a straight line, a roller coaster going down a track, or a projectile motion like throwing a ball. It is also used in engineering and design, such as calculating the trajectory of a rocket or designing a conveyor belt system.

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