The problem in the image states:

• A car has an initial velocity $u = 2 \, \text{ms}^{-1}$.
• The car's acceleration is given by $a(t) = \frac{5}{2} - \frac{t}{8} \, \text{ms}^{-2}$, where $t$ is time.
• We are asked to calculate the distance traveled during the first four seconds.

To solve this problem, we can use kinematic equations, but since the acceleration is time-dependent, we'll need to integrate it to find velocity and then integrate again to find the distance.

Step-by-Step Solution:

1. Find the velocity:

   • Velocity is the integral of acceleration with respect to time: $v(t) = \int a(t) \, dt = \int \left( \frac{5}{2} - \frac{t}{8} \right) dt$
   • Performing the integration: $v(t) = \frac{5}{2}t - \frac{t^2}{16} + C$
   • To find the constant $C$, we use the initial condition that the initial velocity is $u = 2 \, \text{ms}^{-1}$ at $t = 0$: $v(0) = 2 \quad \Rightarrow \quad C = 2$ So the velocity function is: $v(t) = \frac{5}{2}t - \frac{t^2}{16} + 2$

2. Find the distance:

   • Distance (displacement) is the integral of velocity with respect to time: $s(t) = \int v(t) \, dt = \int \left( \frac{5}{2}t - \frac{t^2}{16} + 2 \right) dt$
   • Performing the integration: $s(t) = \frac{5}{4}t^2 - \frac{t^3}{48} + 2t + D$
   • Since the car starts from the origin, $s(0) = 0$, so the constant $D = 0$. Thus, the distance function is: $s(t) = \frac{5}{4}t^2 - \frac{t^3}{48} + 2t$

3. Calculate the distance at $t = 4$ seconds:

   • Substitute $t = 4$ into the equation for distance: $s(4) = \frac{5}{4}(4)^2 - \frac{(4)^3}{48} + 2(4)$ $s(4) = \frac{5}{4} \times 16 - \frac{64}{48} + 8$ $s(4) = 20 - \frac{4}{3} + 8$ $s(4) = 28 - \frac{4}{3} = \frac{84}{3} - \frac{4}{3} = \frac{80}{3}$ $s(4) \approx 26.67 \, \text{meters}$

So, the car travels approximately 26.67 meters in the first 4 seconds.

Would you like more details on any step or have any questions?

Here are 5 related questions you might find interesting:

1. How would the distance change if the initial velocity was different?
2. How do you calculate the maximum velocity of the car?
3. How does the time-dependent acceleration affect the motion compared to constant acceleration?
4. What would the displacement be if the time was extended to 6 seconds?
5. How would you calculate the average velocity over the 4 seconds?

Tip: In kinematics, integrating the acceleration function gives you velocity, and integrating velocity gives you displacement.