To solve this problem, we’ll break it down into the following steps:

1. Find the time it takes for the toy car to reach the end of the 400 cm ramp.

The distance function is given by:

$d(t) = 35t^2 + 40t$

We want to find the time $t$ when the car reaches 400 cm, so set $d(t) = 400$:

$35t^2 + 40t = 400$

Solve this quadratic equation:

$35t^2 + 40t - 400 = 0$

Divide the equation by 5 to simplify:

$7t^2 + 8t - 80 = 0$

Now, apply the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 7$, $b = 8$, and $c = -80$. Substituting these values:

$t = \frac{-8 \pm \sqrt{8^2 - 4(7)(-80)}}{2(7)}$ $t = \frac{-8 \pm \sqrt{64 + 2240}}{14}$ $t = \frac{-8 \pm \sqrt{2304}}{14}$ $t = \frac{-8 \pm 48}{14}$

Now, we solve for $t$:

$t_1 = \frac{-8 + 48}{14} = \frac{40}{14} \approx 2.86 \, \text{seconds}$ $t_2 = \frac{-8 - 48}{14} = \frac{-56}{14} = -4 \, \text{(not valid, since time cannot be negative)}$

Thus, the time it takes for the car to reach the end of the ramp is approximately $t = 2.86$ seconds.

2. Calculate the average velocity of the toy car.

The average velocity $\bar{v}$ is given by:

$\bar{v} = \frac{\text{Total distance}}{\text{Total time}} = \frac{400 \, \text{cm}}{2.86 \, \text{s}} \approx 139.86 \, \text{cm/s}$

3. Find the distance traveled and velocity for $t = 0, 1, 2, 3$.

Distance at different times:

Use the distance function $d(t) = 35t^2 + 40t$.

• At $t = 0$: $d(0) = 35(0)^2 + 40(0) = 0 \, \text{cm}$

• At $t = 1$: $d(1) = 35(1)^2 + 40(1) = 35 + 40 = 75 \, \text{cm}$

• At $t = 2$: $d(2) = 35(2)^2 + 40(2) = 35(4) + 80 = 140 + 80 = 220 \, \text{cm}$

• At $t = 3$: $d(3) = 35(3)^2 + 40(3) = 35(9) + 120 = 315 + 120 = 435 \, \text{cm}$

Velocity at different times:

Use the velocity function $d'(t) = 70t + 40$.

• At $t = 0$: $d'(0) = 70(0) + 40 = 40 \, \text{cm/s}$

• At $t = 1$: $d'(1) = 70(1) + 40 = 70 + 40 = 110 \, \text{cm/s}$

• At $t = 2$: $d'(2) = 70(2) + 40 = 140 + 40 = 180 \, \text{cm/s}$

• At $t = 3$: $d'(3) = 70(3) + 40 = 210 + 40 = 250 \, \text{cm/s}$

Summary of Results:

• Time to reach 400 cm: $t \approx 2.86 \, \text{seconds}$
• Average velocity: $\bar{v} \approx 139.86 \, \text{cm/s}$
• Distances: $d(0) = 0 \, \text{cm}, d(1) = 75 \, \text{cm}, d(2) = 220 \, \text{cm}, d(3) = 435 \, \text{cm}$
• Velocities: $d'(0) = 40 \, \text{cm/s}, d'(1) = 110 \, \text{cm/s}, d'(2) = 180 \, \text{cm/s}, d'(3) = 250 \, \text{cm/s}$

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Do you want more details or have any further questions?

5 Related Questions:

1. How do we interpret the velocity at each point in time?
2. How would the time change if the ramp were 500 cm long?
3. What happens to the velocity as time increases?
4. Can we use calculus to find the acceleration of the toy car?
5. How does the quadratic formula help solve distance-time problems like this?

Tip:

Remember that the derivative of a position function gives you the velocity, and the second derivative gives the acceleration!