Kinematics chasing problem: Car vs. Motorcycle acceleration question

[jerad908]

Homework Statement

Another physics student ‘borrows’ a sports car for a joy ride and discovers that it can accelerate at a rate of 4.90 m/s2. He decides to test the car by challenging Mr. Horn and his motorcycle. Both start from rest, but the student is so confident in his new ride that he gives Mr. Horn a 1.00 s head start. If Mr. Horn moves with a constant acceleration of 3.50 m/s2 and the student maintains his acceleration of 4.90 m/s2, find:
(a) the time it takes the student to overcome Mr. Horn.
(b) the distance he travels before he catches up with Mr. Horn.
(c) the speed of both vehicles at the instant the student overtakes Mr. Horn.

Relevant Equations

Big 5 equations

Im for some reason getting 1.58 s for time.

I found 1.75 m as the head start distance and then I do d=d so: 2.45t^2 = 1.75t^2 +1.75

but the answer for "a" should be 6.45s...

 

[hmmm27]

Why don't you show your work on this one. Given the ambiguity in the previous, perhaps start by saying if you chose $t_0$ after the "headstart" for part a). Does your answer line up with that of the book if you change that ?

 

[SammyS]

Homework Statement:: Another physics student ‘borrows’ a sports car for a joy ride and discovers that it can accelerate at a rate of 4.90 m/s2. He decides to test the car by challenging Mr. Horn and his motorcycle. Both start from rest, but the student is so confident in his new ride that he gives Mr. Horn a 1.00 s head start. If Mr. Horn moves with a constant acceleration of 3.50 m/s2 and the student maintains his acceleration of 4.90 m/s2, find:
(a) the time it takes the student to overcome Mr. Horn.
(b) the distance he travels before he catches up with Mr. Horn.
(c) the speed of both vehicles at the instant the student overtakes Mr. Horn.
Relevant Equations:: Big 5 equations

I'm for some reason getting 1.58 s for time.

I found 1.75 m as the head start distance and then I do d=d so: 2.45t^2 = 1.75t^2 +1.75

but the answer for "a" should be 6.45s...

I found 1.75 m as the head start distance and then I do d=d so: 2.45t^2 = 1.75t^2 +1.75

Not only does Mr. Horn have a head start distance. He also has a head start velocity which you did not take into account.

 

[kuruman]

Instead of trying to figure out the head distance and velocity, it's easier to choose the appropriate "Big" equation and write the position of each vehicle at any time $t$ as shown by the same clock which starts when the first vehicle starts moving. Then say that there is a specific time $t_c~$, the catch-up time, at which the two vehicles are at the same position. No reason to figure out anything else - just solve for the catch-up time.

 

[DannoXYZ]

Can someone show maths to solve this questions? Thx

 

[berkeman]

Can someone show maths to solve this questions? Thx

No, you've been here long enough to know that we don't give solutions when asked. This thread is old enough that it might be okay to show the solution, but what if you have a new homework assignment that matches it?

Instead, please start a new thread in the HH forums with the problem statement and show your work on the solution. If you're having trouble with the equations, we are happy to help once you show your efforts. Thanks.