## anonymous 3 years ago A bus moves away from rest at a bus stop with an acceleration of 1ms^-2. As the bus stars to move a man who is 4m behind the stop runs with a constant speed after the bus. If he just manages to catch the bus find his speed?

1. anonymous

@Callisto @davian @inkyvoyd

2. Callisto

I don't know if it works. |dw:1346211401321:dw| Let s be the distance travelled by bus, distance travelled by the man = s+4 Time for the bus and man travelled is the same For the bus: s = ut + (1/2) at^2, where u = 0ms^-1, a = 1ms^-2 For the man, s + 4 = vt No... it doesn't work. I need one more equation :(

3. anonymous

yup..i tried to buy the answer is $2\sqrt{2} ms^{-1}$

4. Callisto

I know it doesn't make sense... s = ut + (1/2) at^2, where u = 0ms^-1, a = 1ms^-2 So, s = 0 + (1/2)(1)(t^2) s = t^2 / 2 -(1) s + 4 = vt -(2) (2)- (1), what do you get?

5. anonymous

no the books answer is tht, i tried nd i dnt get it o.O

6. Callisto

What do you get for (2) - (1) ???

7. Callisto

I meant my way to get the answer doesn't make sense..

8. anonymous

so theres no solution?

9. Callisto

No no no,... You don't even answer my question :| s = t^2 / 2 -(1) s + 4 = vt -(2) (2) - (1): 4 = vt - t^2/2 Agree?

10. anonymous

yes i agree:)

11. Callisto

Now, rearrange it. -t^2/2 + vt - 4 = 0 Quadratic equation. I take an assumption here: there is only one intercept for the x-axis, which means the man can only catch the bus at one time one intercept: delta = 0 That is v^2 - 4(-1/2)(-4) = 0 Solve v.

12. Callisto

I know it doesn't make sense at all...

13. anonymous

it does:) $\sqrt{8}--->2\sqrt{2}$

14. Callisto

I don't think it is the normal way to work out the solution... That's why I said this method doesn't make sense. :(

15. anonymous

can u explain ur assumption?

16. anonymous

i understand b^2 - 4ac = 0

17. anonymous

thanks alot :)

18. Callisto

That's the most nonsense point. Since the bus is accelerating, the man doesn't run faster to catch the bus, when the man misses that chance, the bus would be further away from him and it's not possible to catch the bus if he doesn't accelerate

19. anonymous

yes..evn i thought the same thng

20. Callisto

But there is a bug? :|

21. anonymous

wt u mean?

22. Callisto

Hmm.. I don't know :|

23. anonymous

anyway thnks once again :)

24. Callisto

Welcome. but better wait if some others can give a more reasonable way to do this question! I'm sorry!

25. ash2326

Let the bus covers a distance x, while the man runs so the man has to run for a distance of (4+x) meters Let this time be t and the constant speed of man be c $(4+x)=c\times t$ now for the bus $S=ut+\frac 12 at^2$ $x=0+\frac 12 1 t^2$ so $t^2=2x$ we have the other equation as $4+x=ct$ x=t^2/2 so we have $8+t^2=2ct$ rearranging we get $t^2-2ct+8=0$ for a solution to exist $b^2-4ac \ge0$

26. anonymous

This defines the movement of the bus:$x_b=x_{0b}+v_{0b}t+\frac{1}{2}a_bt^2$And this defines the movement of the man:$x_m=x_{0m}+v_{0m}t+\frac{1}{2}a_mt^2$We want to see what the velocity of the man is when it's x position is the same as the bus', so let's set their positions equal to eachother:$x_{0b}+v_{0b}t+\frac{1}{2}a_bt^2=x_{0m}+v_{0m}t+\frac{1}{2}a_mt^2$We can express the bus' initial position in terms of the man's initial position:$x_{0b}=x_{0m}+4m$We also know that the acceleration of the man is 0 (constant speed) and that the initial velocity of the bus is 0. This means we can cancel some things out in our equation:$(x_0m+4m)+\frac{1}{2}a_bt^2=x_0m+v_{0m}t$x0 of the man cancels:$4m+\frac{1}{2}a_bt^2=v_{0m}t$Now we are down to 2 unknown variables. We need to express one of the two unknowns (v_0m and t) in terms of eachother, so let's look at another equation:$v_{fm}=v_{0m}+a_mt \rightarrow v_{fm}=v_{0m}$We know that if the man JUST catches the bus, their velocities will be equal at the time that the man catches it, so we can assume that:$v_{fm}=v_{fb}$The final velocity of the bus can be defined as:$v_{fb}=v_{0b}+a_bt \rightarrow v_{fb}=a_bt$Since final velocities are equal:$v_{fb}=a_bt \rightarrow v_{fm}=a_bt$Let's plug this back into our equation:$4m+\frac{1}{2}a_bt^2=v_{0m}t \rightarrow 4m+\frac{1}{2}a_bt^2=(a_bt)t$$4m+\frac{1}{2}a_bt^2=a_bt^2$$4m=\frac{1}{2}a_bt^2 \rightarrow t=\sqrt{\frac{8m}{a_b}}$$t=\sqrt{\frac{8m}{+1\frac{m}{s^2}}}=\sqrt{8}s$We can now plug this back into one of our old position equations to find v_0m (and therefore v_fm):$4m+\frac{1}{2}a_bt^2=v_{0m}t \rightarrow v_{0m}=\frac{4m+\frac{1}{2}a_bt^2}{t}=\frac{4m}{t}+\frac{1}{2}a_bt$$v_{0m}=v_{fm}=\frac{4m}{t}+\frac{1}{2}a_bt \rightarrow v_{fm}=\frac{4m}{\sqrt{8}s}+\frac{1}{2}(1\frac{m}{s^2})(\sqrt{8}s)$$v_{fm}=+2.83\frac{m}{s}$Which matches up with other answers given.