Girl chasing boy at constant then accelerated speed

[CathyLou]

Hi.

Could someone please help me to do part (b) of the following question? I would really appreciate any help as I am really stuck.

A boy is walking at a constant speed of 1.8 m/s along a straight road. He passes a telephone booth where his sister is making a telephone call. His sister takes 30 s to complete the call and then sets off in pursuit of the boy. She accelerates uniformly from rest at 3 m/s^2 until she is running at a speed of 9 m/s. She maintains this constant speed until she reaches her brother.

(a) On the same diagram sketch the speed-time graphs for the boy and his sister.

I did this part.

(b) Calculate the time taken by the girl to reach her brother.

Thank you.

Cathy

 

[malawi_glenn]

What have you tried so far?

I'll give you a hint: what does the area of the speed-time graph represent?

 

[CathyLou]

What have you tried so far?

I'll give you a hint: what does the area of the speed-time graph represent?

Thanks for replying.

It represents the distance travelled.

Cathy

 

[malawi_glenn]

So what is meant by "girl to reach her brother" ?
Well it means that they have traveled same distance, i.e find the value of time t, when the area of girl = area of boy. Do you know how to go from lines on a graph to functions and do integration?

I have no time to draw pictures and so on, I am quite sure that you know how this works.

If not, just use the fact that the boy will have area: 1.8*t and the find the time for the girl to reach 9 m/s with the acceleration given and then just continue.

 

[CathyLou]

So what is meant by "girl to reach her brother" ?
Well it means that they have traveled same distance, i.e find the value of time t, when the area of girl = area of boy. Do you know how to go from lines on a graph to functions and do integration?

I have no time to draw pictures and so on, I am quite sure that you know how this works.

If not, just use the fact that the boy will have area: 1.8*t and the find the time for the girl to reach 9 m/s with the acceleration given and then just continue.

Thanks so much for your help!

Cathy