I with a physics problem about acceleration and velocity

[meandyou741]

Homework Statement

Jane is riding her new bike at a constant velocity of 3 m/s when she goes past Tom. Tom notices that the road will come to an end with a 30m drop. Tom grabs his bike and attempt to catch up to Jane. Jane has a 20 sec head start. Tom is going to ride at a constant acceleraion of 1/2 m/s^2 for 20 sec then at a constant velocity until he rescues Jane or Jane rides to disaster. when Tom has been traveling for 8 sec Jane see that he is trying to catch up and accelereated at a rate of 5/12 m/s^2 for 12 sec then continuing at a constant velocity.

A) if the road is 375 m. long will Tom rescue her?
B) is jane flies over the edge, how many seconds before she hits the ground?
C) if Tom does catch up to Jane, how long after Tom starts will he catch up to Jane?

Homework Equations

V= Vo+at
V^2= Vo^2+2ax
D=vt
A=Vt

The Attempt at a Solution

i found Toms velocity using a=vt and got 10m/s
jane has been traveling at 3ms for 28 sec.
janes velocity after she starts to accelerate , using a=vt, is 5 m/s
Jane traveled 84 meters at a constant velocity of 3m/s and.
Tom went a total of 200 meters while accelerating.
Jane wen 60 more meters after she started to accelerate.

i don't really know what to do from here. how do i start?

 

[Kurdt]

You need to work out the distance both go before they meet. You will need other kinematic equations.

https://www.physicsforums.com/showthread.php?t=110015

 

[meandyou741]

Would I use the X =Vo+ 1/2 a t^2
And. How would I start ? Can u help me begin?

 

[henryc09]

For a) you could try to work out the position of tom and jane (relative to tom's start position) as a function of time and then find how long it takes jane to reach the end (375m) and compare this to how long it takes tom.

for (b) just consider the vertical velocity, which is initially 0 and use the constant acceleration formulae.

 

[henryc09]

actually looking at the problem the best thing to do I think is to evaluate how far they've both moved from tom's initial position after 20s (because at this point they're both traveling with constant velocity). Then work out their velocities at this time, and from that you can easily work out how long it will take each to reach the end of the road.

 

[meandyou741]

Ok so I got thta for tom it takes 75 sec. ( doing 375= .5*10 T) and for jane it would take 88.75 sec. ( doing 375= .5*(3+5)T but then I subtracted 20 from jane's time because she had a head start and got 68.5 menaing that Tom does not catch up t her in time.

Then I did ( -30=.5(9:8)T^2 ) and got 5.oq sec for her to hit the water.

Can u confirm I did it correctly?

 

[Kurdt]

You're not taking into consideration the different phases of motion. Tom isn't accelerating all of the time and neither is Jane. Deconstruct the problem into the various distinct periods of motion and analyse from there.

 

[verty]

To make this easier, draw a timeline with all the relevant points on it, and write in the distances that Tom and Jane have traveled at each point. Imagine that you are watching an action replay. Jane is getting ahead, then Tom starts to catch up, then Jane accelerates. Get a picture of just where they are at all times.