Finding Distance Traveled with Varying Acceleration

In summary, the motor drives a body along a straight line with a constant force. The power P delivered by the motor must vary with time according to v.P=Fv, where F is constant, so P must vary with time according to v.
  • #1
vipulgoyal
26
0

Homework Statement



a motor drives a body along a straight line with a constant force. The power P delivered by the motor must vary with time t according to?
where t is time and conditions are ideal

Homework Equations



P = F v
x=vt

The Attempt at a Solution


according to me answer comes as t-1 but as you know that's not the answer is t just that
 
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  • #2
Your question is unclear.
 
  • #3
vipulgoyal said:
a motor drives a body along a straight line with a constant force. The power P delivered by the motor must vary with time t according to?
where t is time and conditions are ideal

P=Fv, where F is constant, so P must vary with time according to v.
 
  • #4
I like Serena said:
P=Fv, where F is constant, so P must vary with time according to v.

i felt the same when i read the question
 
  • #5
vipulgoyal said:

Homework Statement



a point initially at rest moves along x-axis its accelaration varies with time as a= 6t+5 if it starts from the origin the distance covered in 2s is

Homework Equations




s= ut +1/2at2

The Attempt at a Solution



ijus tried put the values of different variable in the equation but the answer doesnot come that way is their any prob with it??
 
  • #6
vipulgoyal said:
i felt the same when i read the question

Actually we can say a bit more.
Under ideal conditions I assume the friction is zero.

That means:

[tex]a = \frac F m[/tex]

[tex]v = a t = \frac F m t[/tex]

[tex]P = F v = \frac {F^2} m t[/tex]

So P varies linearly with t.
 
  • #7
vipulgoyal said:

Homework Statement



a point initially at rest moves along x-axis its accelaration varies with time as a= 6t+5 if it starts from the origin the distance covered in 2s is

Homework Equations



s= ut +1/2at2

The Attempt at a Solution



ijus tried put the values of different variable in the equation but the answer does not come that way is their any prob with it??

Your equation only holds when a is constant.

You need the following equations:

[tex]\begin{equation}v(t) = \int_0^t a(t) dt \label{eq:a_integral}\end{equation}
\begin{equation}s(t) = \int_0^t v(t) dt \label{eq:v_integral}\end{equation}[/tex]
 

1. What is the definition of work?

In physics, work is defined as the product of force and displacement, where a force is applied to an object causing it to move a certain distance in the direction of the force.

2. How is work measured and what is the unit of measurement?

Work is measured in joules (J) and is calculated by multiplying the force applied to an object by the distance it moves in the direction of the force.

3. What is the relationship between work and energy?

Work and energy are closely related concepts in physics. Work is the transfer of energy from one object to another, or the transformation of energy from one form to another. The amount of work done on an object is equal to the change in its energy.

4. Can work be negative?

Yes, work can be negative. This occurs when the force applied to an object is in the opposite direction of its displacement. In this case, work is considered to be done against the object, and the work done is negative.

5. How is power related to work and energy?

Power is the rate at which work is done or energy is transferred. It is measured in watts (W) and is calculated by dividing the amount of work by the time it takes to do the work. This means that a higher power output indicates a faster rate of work or energy transfer.

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