1. Which of the graphs correctly displays the positions of the motorcycle and car as functions of time?
2. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1.
3. How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2)?

Question

1. Which of the graphs correctly displays the positions of the motorcycle and car as functions of time?
2. How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1.
3. How far does the motorcycle travel from the moment it starts to accelerate (at time t1) until it catches up with the car (at time t2)?

mony01 · Answer

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle are both traveling at the same speed of 23.0m/s , and the distance between them is 50.0m . After t1 = 3.00s , the motorcycle starts to accelerate at a rate of 6.00m/s2 . The motorcycle catches up with the car at some time t2.

mony01 · Answer

attachment

ahsome · Answer

1) Graph A

ahsome · Answer

2) T1=3 Seconds, T2=6 Seconds
Time difference+
$$T1-T2$$
$$6-3=3$$

ahsome · Answer

3)$$Distance=Speed*Time$$
$$D=6*3$$
$$D=18m$$

ahsome · Answer

Do you understand @mony01?

mony01 · Answer

hold on for number 2 so you're saying that T2 is 6?

mony01 · Answer

I don't think thats right because the answer you got was wrong

ahsome · Answer

Ok then. That means we need to find T2, not assuming that T2 is T1*2

mony01 · Answer

how we do that?

ahsome · Answer

You know that he the motorbike closes the gap between the car at 6.00ms^2 acceleration

ahsome · Answer

The gap is 50 metres

mony01 · Answer

so we divide 6 by 50?

ahsome · Answer

@hartnn We need your help

hartnn · Answer

s = ut +1/2 at^2 

s=50
u= 23
a=6

find t
that will be t2

mony01 · Answer

how did you get this equation?

ahsome · Answer

This equation is one of the basic Physics equation

mony01 · Answer

Can someone verify if T2 is 50/3?

ahsome · Answer

Don't you have the answers with you?

mony01 · Answer

no

mony01 · Answer

@hartnn

ahsome · Answer

s = ut +1/2 at^2 s=50 u= 23 a=6
$$s=ut+\frac{ 1 }{ 2 }at^2$$
$$50=23t+\frac{ 1 }{ 2 }6t^2$$
$$50=23t+3t^2$$
$$3t^2+23t-50=0$$

mony01 · Answer

thats the part i got stuck on how i solve for t?

ahsome · Answer

Use the quadratic equation
$$\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }$$

ahsome · Answer

$$3t^2+23t-50$$
The A value is 3, B value is 23, C value is -50

ahsome · Answer

Substitute into the equation:
$$\frac{-b\pm\sqrt{b^2-4ac}} {2a}$$
$$\frac{-23\pm\sqrt{23^2-4*3*-50}} {2*3}$$

ahsome · Answer

Do you understand @mony01?

mony01 · Answer

yea i got 1.767 is it right?

ahsome · Answer

I haven't done the sum myself, but If you did it right, then yeah :)

mony01 · Answer

do i still subtract 3

ahsome · Answer

Yes, you have to

ahsome · Answer

Remember, this equation has 2 answers.

mony01 · Answer

yea one is negative do i use it?

ahsome · Answer

No, never use negative number. Use positive

ahsome · Answer

Only use negative when it isn't real-life, like in a graph

mony01 · Answer

@Ahsome thanks I found a solution without having to use the quadratic formula thank you anyways

ahsome · Answer

Oh, Good job :D (How did you do it?)