1. ## Mechanics, two questions

Q1-A truck with initial velocity 6 m/s^-1 brakes and comes to rest. At time t seconds after the brakes are applied with acceleration a m/s^-2 where a=-3t. This formula applies until the truck halts.

1: Find time taken for the truck to halt.
2:Find distance traveled whilst decelerating.
3:Find greatest deceleration of truck?

Q2-A force of (36-t^2) newtons acts at time t seconds on a particle of mass 1kg. When t=0 the particle has velocity 2m/s^-1 in the direction of the force. Find the velocity of the particle when t=6 and find the distance travelled between t=0 and t=6.

2. $a = -3t$

$
v(t) = \int -3t \, dt = -\frac{3t^2}{2} + C
$

$v(0) = 6 \, m/s$ ... $6 = 0 + C$

$
v(t) = -\frac{3t^2}{2} + 6
$

1) set $v(t) = 0$ and solve for t

2) $d = \int_0^{t_1} v(t) dt$, where $t_1$ is the time found in question (1).

3) think about it ... when will -3t have its greatest magnitude?

3. Originally Posted by skeeter
$a = -3t$

$
v(t) = \int -3t \, dt = -\frac{3t^2}{2} + C
$

$v(0) = 6 \, m/s$ ... $6 = 0 + C$

$
v(t) = -\frac{3t^2}{2} + 6
$

1) set $v(t) = 0$ and solve for t

2) $d = \int_0^{t_1} v(t) dt$, where $t_1$ is the time found in question (1).

3) think about it ... when will -3t have its greatest magnitude?
Thanks for your help. I have done parts one and two and got t=2 and 8m for distance travelled, in relation to magnitude, I am still stuck .

Can anybody help me out with Question 2? Im so puzzled

4. $
a(t) = -3t
$

$
a(0) = 0
$

$
a(1) = -3
$

$
a(2) = -6
$

what is the greatest magnitude for $a$ ?

btw, magnitude of $a$ would be $|a|$