Finding the Magnitude of the Force Acting on an Object

[correction_tape]

Homework Statement

The velocity (m/s) of an object at time (t) is given by v = -4.0 + 2.0t.

(1) Calculate the magnitude of the force F acting on the object of mass m = 4.0kg.
(2) Calculate the kinetic energy (K) of the object of mass m = 4.0kg at time t = 3.0s.
(3) Setting the initial position x(subscript 0) = 0.0m of the object at time t = 0.0s, calculate the time (t>0.0s) at the position x=21m of the object.

Relevant Equations

F = ma
a = dv/dt

So far, I've only attempted the first number:

F = (4kg) [(-4+2t)-0]/t

?

Been thinking how to cancel out the time, but I really have no idea how to proceed. Was I supposed to cancel it even? Can someone please help and tell me how you would answer this? (And maybe show a solution as well so I can make sense of what to do when I encounter these problems again)

By the way, this is not homework. I'm just studying from exam papers online.

 

[Orodruin]

Do you understand how derivatives work? You have divided velocity by t, not taken its derivative with respect to t.

 

[correction_tape]

Do you understand how derivatives work? You have divided velocity by t, not taken its derivative with respect yo t.

Sorry, not very much. How do you do that?

 

[Orodruin]

A derivative is a measure of how fast something changes with a variable. In this case how velocity changes with time, which by definition is acceleration. Also in this case, velocity changes linearly with time so acceleration is constant.

Derivatives is generally something you will want to study before trying to dig deeper into kinematics (or any subfield of physics really). This is an entire subject by itself so it is not really suitable for a PF post.

Are you familiar with the SUVAT equations?

 

[Delta2]

This can be solved without derivatives but you must know SUVAT equations and the related kinematics theory.
Also Newton's 2nd law is needed but you have that in the list of related equations in the OP.

Derivatives is generally something you will want to study before trying to dig deeper into kinematics (or any subfield of physics really)

Yes , derivative and integrals, differential and integral calculus in general is almost absolutely basic stuff (along with linear algebra) for the way physics are being taught today.

 

[BvU]

Sorry, not very much. How do you do that?

In simpler language, your $a=dv/dt$ means $a$ is the slope of the graph of $v$ as a function of $t$.
Since, in your case, this graph is straight line, the expression for $a$ is $$a= { v(t+\Delta t) - v(t)\over \Delta t} = { -4+2( t+ \Delta t ) - ( -4+2t ) \over \Delta t}= ...$$

$\$

 

[Steve4Physics]

Hi @correction_tape. Welcome to PF. I’m guessing you are working at an introductory/shool level, so differentiating (i.e. using calculus) may not be appropriate.

The velocity (in m/s) of an object at time t (in s) is given by v(t) = -4.0 + 2.0t.

Look at the equation carefully. Can you tell how much v changes every time t increases by 1 second?

That’s the rate of change of v, i.e. the acceleration.

 

[Mister T]

v = -4.0 + 2.0t.

Compare this to the equation $v=v_o+at$.