Calculus 1 : Solving for Time

Study concepts, example questions & explanations for Calculus 1

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Example Questions

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Example Question #31 : Solving For Time

A particle is accelerating at a rate given by the expression , where t is the time in seconds after the particle passes a gate inside of the particle accelerator. Two seconds after passing the gate, a radar detects that the particle was moving at a velocity of 50 m/s. The particle will continue to accelerate until it hits a lead wall found approximately 7,000 meters after passing the gate.

Find an expression in terms of t that gives the amount of time in seconds it takes for the particle to reach the wall after passing the gate.

Possible Answers:

Correct answer:

Explanation:

The acceleration of the particle is given by the follwoing formula: . In order to find the velocity of the particle, this need to be indefinetely integrated.

When taking an integral, you can seperate the integral into the sum of the integral of each part.

Therefore, the expression for the velocity of the particle is determined to be:

.

In order to find the constant C, you must use the conditions provided to you. The radar detects the velocity of the particle to be equal to 50 m/s two seconds after the particle passes the gate. Use this information to find the constant C.

Now that the expression for velocity has been determined, the position function can be found by integrating the velocity function. The position function is used to find how long it takes for the particle to travel 7000 meters.

Once again, this integral can be seperated into a sum of the integral of each part:

 

Since the start of the 7000 meters begins at the gate, the initial position of the particle is zero. Therefore, the constant C is equal to zero. This gives a position function for the particle:

By setting this equal to the total distance traveled, the time t that it takes to reach 7000 meters can be determined.

or

 

 

 

 

 

 

Example Question #32 : Solving For Time

A particle's velocity is modeled (in feet per second) by the equation: 

At , the particle is at rest.

At what time is the particle's acceleration equal to 578 ft/s/s?

Possible Answers:

Correct answer:

Explanation:

For this question, we must first determine what the accleration function is by taking the derivative of the velocity equation:

Recall the power rule to differentiate.

The power rule states,

The equation for the acceleration of the particle, then, is:

With this equation in hand, we just have to solve for t when :

We are only interested in the positive solution to this problem, since the particle is not in motion before t=0.

Example Question #33 : Solving For Time

Find how long it takes for a projectile starting at  to hit a critical point if its height  is given by:

Possible Answers:

Correct answer:

Explanation:

To solve this, we have to take the first derivative and set it equal to zero:

Remember that derivative of  and derivative of 

Moving  to the other side,

We now get that:

From here, we can see that this equation will never equal zero since 

 will never have a solution since 

The correct answer is 

Example Question #34 : Solving For Time

There are equations in physics that can tell you about the position and velocity of an object as a function of time.  If we were to graph the equation of the relationship between the object's height and time it would be a quadratic while graphing the velocity vs time would be linear (since velocity is the derivative of position in respect to time.  The equation of the height of the projectile with respect to time is the following where  is the initial height,  is the acceleration, and  is the initial velocity.  

Say you throw a ball straight up in the air at an initial velocity of 20m/s.  At approximately what time does the ball reach it's highest height?  Use  as the approximation for the acceleration due to gravity.  

Possible Answers:

Not enough information given.

Correct answer:

Explanation:

Given that the relationship between height and time is a parabola, we note that the greatest height will occur when the first derivative is zero.

Given the equation,

 

The derivative is 

 

which we then set equal to zero and plug in our known values.

We get the following: 

.  

Then solving for time, t, gives us 2.04s.

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