1. SUVAT Question

A particle is moving along the x-axis with constant deceleration 5ms-2. At time=0, the particle passes through the origin o with velocity 12ms-1 in the positive direction. At time t secounds the particle passes through the point A with x-coordinates 8. Find the values of t.

Any help?

I know the formula that is used is s=ut+1/2at^2 but I keep getting the wrong answer.

2. Originally Posted by Mist
A particle is moving along the x-axis with constant deceleration 5ms-2. At time=0, the particle passes through the origin o with velocity 12ms-1 in the positive direction. At time t secounds the particle passes through the point A with x-coordinates 8. Find the values of t.

Any help?

I know the formula that is used is s=ut+1/2at^2 but I keep getting the wrong answer.
s = 8 (because we've gone from 0 to 8 on the line)
t = t
u = 12
a = -5 (the minus sign is important)
$\displaystyle s = ut+ \frac{1}{2}at^2$

$\displaystyle 8 = 12t - \frac{5}{2}t^2$

Multiply through by 2 to clear the fraction and rearrange into the form $\displaystyle ax^2+bx+c=0$

$\displaystyle 5t^2-24t+16=0$

Solve using your favourite method, this one does factorise

$\displaystyle (5t-4)(t-4) = 0 \: \: \therefore \: t = \frac{4}{5} \: \text{ or } \: t = 4$

To find the value find v at the given time

$\displaystyle v_1 = u+at_1 =12 - 4 = 8$

$\displaystyle v_2 = u+at_2 = 12-20 = -8$

Since $\displaystyle v_2 < 0$ it implies that the particle is now moving in the negative direction, back towards the origin. Therefore it follows that t = 0.8 is correct.

oops, just saw that the question asks for the values of t (implying more than one solution) so if you quote both you should be correct and can disregard the stuff about v