# Thread: Velocity and Acceleration

1. ## Velocity and Acceleration

The distance traveled by a car in a time t seconds afer the brakes are applied is given by:

$y= 10t-1.8t^2$

Where y is the distance in metres and t is time in seconds.

Determine equations for velocity and acceleration of the car.

What is the velocity at the instant the brakes are applied?

How many seconds does it take the car to stop?

-----------------------------------

$10t$

$1.8t^2 = 3.24t$

$10t - 3.24t = 6.76$

$Time = 6.76 Seconds?$

Whats the next step? and is that correct so far? help welcomed please.

I know that $V= dx/dt$

2. Originally Posted by ForgottenMemorie
The distance traveled by a car in a time t seconds afer the brakes are applied is given by:

$y= 10t-1.8t^2$

Where y is the distance in metres and t is time in seconds.

Determine equations for velocity and acceleration of the car.

What is the velocity at the instant the brakes are applied?

How many seconds does it take the car to stop?

-----------------------------------

$10t$

$1.8t^2 = 3.24t$

$10t - 3.24t = 6.76$

$Time = 6.76 Seconds?$

Whats the next step? and is that correct so far? help welcomed please.

I know that $V= dx/dt$

$y= 10t-1.8t^2$
$dy/dt=10-3.6t$
$v=10-3.6t$
$dv/dt=-3.6$
$a=-3.6$

$v=10-3.6t ,t=0$
$v=10$

$v=10-3.6t$
$0=10-3.6t$
$t=2.778$

3. Okies that really does help. Thanks for thte responce.