Two cars,a and b travel in a straight line.The distance of a from the starting point is given by x=(4m/s)t+(1m\s^2)t^2;the distance of b from starting point is y=(2m/s^2)t^2+(2m/s^3)t^2;

(a)At what times are the cars at the same point?
(b)At what times is the velocity of b relative to a zero?

the formula for y (distance) is bogus

m/s^3 * t^2 = m/s

the units don't work out right.

It could be that you meant
x = 4t + t^2
y = 2t + 2t^2

If so, then

(a) set x=y and solve for t
(b) set the speeds equal and solve for t

To find the times when the two cars are at the same point, we need to set their distances equal to each other and solve for t. Let's start with part (a).

(a) Setting the distances equal to each other:
x = y

(4m/s)t + (1m/s^2)t^2 = (2m/s^2)t^2 + (2m/s^3)t^3

Next, we can simplify the equation by moving all the terms to one side:
(2m/s^2)t^2 + (2m/s^3)t^3 - (4m/s)t - (1m/s^2)t^2 = 0

Combine like terms:
(1m/s^3)t^3 + (1m/s^2 - 4m/s)t^2 - (4m/s)t = 0

Now, we have a cubic equation. Since it doesn't factor nicely, we can use numerical methods like Newton-Raphson or simply graph the function to estimate the solutions.

For part (b), we need to find the times at which the velocity of car b relative to car a is zero. The velocity of b relative to a can be found by taking the derivative of the distance equation for car b with respect to time (t).

Let's find the derivative of y with respect to t:

dy/dt = 2(2m/s^2)t + 2(2m/s^3)(2t)

Simplifying:
dy/dt = 4(2m/s^2)t + 4(2m/s^3)t^2

Now, we can set the velocity of b relative to a equal to zero:

dy/dt = 0

4(2m/s^2)t + 4(2m/s^3)t^2 = 0

4(2m/s^2)t + 4(2m/s^3)t^2 = 0

Divide both sides by 4 and simplify:
(2m/s^2)t + (2m/s^3)t^2 = 0

Again, we have a quadratic equation this time. We can solve this equation using either factoring or the quadratic formula.

Note that in both parts (a) and (b), we need to solve the equations to find the values of t at which the conditions are met.