Kinematics no number solution question issue

[mrjoe2]

To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is "burning out" at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.)

You drive at a constant speed of v0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, a . Let the time at which the dragster starts to accelerate be t=0 .

What is tmax, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?




i believe the only equation that would be required is v=d/t. it is the only one that i used but still got the incorrect answer.



when i worked it out i thought the answer would be delta d/vo, but it wasnt the correct answer. it seemed perfect for an expression as an answer to the maximum time it would take for the car to collide with the dragster because when the dragster reaches vo which is the speed of the car, that's the last possible time they could possible collide right?

 

[Hootenanny]

To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is "burning out" at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.)

You drive at a constant speed of v0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, a . Let the time at which the dragster starts to accelerate be t=0 .

What is tmax, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?

i believe the only equation that would be required is v=d/t. it is the only one that i used but still got the incorrect answer.
when i worked it out i thought the answer would be delta d/vo, but it wasnt the correct answer. it seemed perfect for an expression as an answer to the maximum time it would take for the car to collide with the dragster because when the dragster reaches vo which is the speed of the car, that's the last possible time they could possible collide right?

The equation v=d/t is only valid for constant velocity and not when an object is accelerating. I would suggest that the problem is best tackled using kinematic equations.

 

[mrjoe2]

The equation v=d/t is only valid for constant velocity and not when an object is accelerating. I would suggest that the problem is best tackled using kinematic equations.

i understand that, however, the question is asking for a specific point in time, which is at a certain velocity, and not during the time the dragster is accelerating. ( i am first year university physics and it is quite hard for me)

 

[Hootenanny]

i understand that, however, the question is asking for a specific point in time, which is at a certain velocity, and not during the time the dragster is accelerating. ( i am first year university physics and it is quite hard for me)

Whilst that is true, v=d/t is not valid at a specific point in time, the equation requires a change in distance and a change in time and should really be written:

$$v = \frac{\Delta x}{\Delta t}$$

 

[mrjoe2]

Whilst that is true, v=d/t is not valid at a specific point in time, the equation requires a change in distance and a change in time and should really be written:

$$v = \frac{\Delta x}{\Delta t}$$

that was my original solution. i rearranged v = dd/dt to get dtmax = dd/vo and i thought that would be the solution. but it was incorrect. is there anything that you can do to lead me to a correct expression that would be a solution to this question?

 

[Hootenanny]

that was my original solution. i rearranged v = dd/dt to get dtmax = dd/vo and i thought that would be the solution. but it was incorrect. is there anything that you can do to lead me to a correct expression that would be a solution to this question?

As I've said previously, your solution was incorrect simply because that equation is not valid for accelerating objects, nor at an instant in time. You need to use kinematic (SUVAT) equations to solve this problem.

 

[mrjoe2]

As I've said previously, your solution was incorrect simply because that equation is not valid for accelerating objects, nor at an instant in time. You need to use kinematic (SUVAT) equations to solve this problem.

just to make clear. there is no solid number answer to this question right? it is an expression?

 

[Hootenanny]

just to make clear. there is no solid number answer to this question right? it is an expression?

Correct.