Answer
a) 300ft
b) 60 ft/sec
Work Step by Step
Let the acceleration be a, velocity be v and displacement be s
a) Let the truck have the velocity $v_{t}$= 30
$s_{t}=\int v_{t}$ dt
=30t+c where c is an arbitrary constant
Taking the traffic light to be the reference point,
$s_{t}$=30t
Let the car have acceleration $a_{c}$=6
$v_{c}=\int a_{c}$ dt
=6t+c where c is an arbitrary constant
Since the car started from rest,
$v_{c}$=6t
$s_{c}=\int v_{c}$ dt
=3t$^{2}$+c where c is an arbitrary constant
Taking the traffic light to be the reference point,
$s_{c}$=3t$^{2}$
To find the time when the car overtakes the truck, set $s_{c}$=$s_{t}$
3t$^{2}$=30t
3t(t-10)=0
t=0 or t=10
The car will pass the truck at time t=10 seconds.
Now, we find the distance by plugging back into the equation:
$s_{c}$=3t$^{2}$
$s_{c}$=3(10)$^{2}$
$\boxed{s_{c}=300\text{ ft}}$
b) At t=10, velocity of car $v_{c}$= 6(10)
=60 m s$^{-1}$
