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Sodium · Sep 18, 2009, 05:49 PM

Hello everyone!
Just going over my physics review, and realized I have no idea about one of these questions, please help and explain what you are doing, THANKS:)

A crazy transport truck who was driving 130km/h behind mr. Berk spots mr berks VW jetta at a distance of 75 m from his bumper. He hits the brakes.

a) if his reaction time is .8s how far does he travel before the brakes are engaged

b) the truck begins decellerating at 15m/s squared fish tailing widly. Does he hit mr berks car?

Please explain!

And another one...

A ball is launched vertically in the air with a initial velocity or 23m/s [up]. How high with the ball travel up from the point where it was launched (hint the ball will decelerate at the same rate it would accelerate had it been dropped)

Ugh!

Unknown008 · Sep 19, 2009, 01:11 AM

1. a) The truck is travelling 130km/h. In 0.8 s, he makes how much distance?

Use $Velocity=\frac{Distance}{Time}$

Be careful of the units!

b) Ok, you have the initial speed, the decceleration. Use $v^2 = u^2+2as$

v is the final speed (0 km/h)
u is the initial speed (130 km/h)
a is the acceleration. Since it deccelerates, the value is -15m/s^2
s is the displacement.

Solve for s. If s is larger than the remaining distance (75 m - thinking distance), then, the collide. Be careful of the units once again!

2. Use again $v^2 = u^2+2as$

When the ball reaches its highest point, its velocity will be 0. Solve for s.

I hope it helped! :)

Sodium · Sep 20, 2009, 09:09 PM

I understand 1 a now!
But the other two I am still confused on, I don't know those formulas, so it throws me off.

Unknown008 · Sep 20, 2009, 11:12 PM

Ok, I'll do the derivative of the formula $v^2=u^2+2as$

You know that $v=u+at$ ?

That's from the definition of acceleration: $a = \frac{v-u}{t}$

Now, you also know that $s = \frac{v+u}{2} t$ (Total Displacement = Average Speed x Total Time[/math]

In the first formula, make t the subject;

$t = \frac{v-u}{a}$

Substitute t in the second formula;

$s = \frac{v+u}{2} \frac{v-u}{a}$

Expand;

$s = \frac{(v+u)(v-u)}{2a}$

$s = \frac{v^2-u^2}{2a}$

And finally simplify;

$2as = v^2-u^2$

$2as+u^2 = v^2$

$v^2 =u^2 + 2as$

That formula is very important, because it allows to to solve certain problems when you don't have the time.

Okay? :)