A lorry accelerates uniformly in a straight line with acceleration of 4ms^-2 and covers a distance of 250 m in…

1. Analyze the Question

We are given the acceleration ($a = 4 \, \text{ms}^{-2}$) and the distance covered in the first 10 seconds ($s_1 = 250 \, \text{m}$). We need to find the distance covered in the next 10 seconds.

2. Explain the Concept

To solve this, we use the equations of motion for uniform acceleration:

• $s = ut + \frac{1}{2}at^2$
• $v = u + at$

Where:

• $u$ = initial velocity
• $v$ = final velocity
• $a$ = acceleration
• $t$ = time
• $s$ = distance

3. Show the Solution

Step 1: Find the initial velocity ($u$) using the first interval. Given: $s_1 = 250 \, \text{m}$, $t_1 = 10 \, \text{s}$, $a = 4 \, \text{ms}^{-2}$. Using $s = ut + \frac{1}{2}at^2$: $250 = u(10) + \frac{1}{2}(4)(10)^2$ $250 = 10u + 2(100)$ $250 = 10u + 200$ $50 = 10u$ $u = 5 \, \text{ms}^{-1}$

Step 2: Find the velocity at the end of the first 10 seconds ($v_1$). This will be the initial velocity for the next 10 seconds. $v_1 = u + at_1$ $v_1 = 5 + (4)(10)$ $v_1 = 5 + 40 = 45 \, \text{ms}^{-1}$

Step 3: Calculate the distance covered in the next 10 seconds ($s_2$). For this interval, the initial velocity is $u_2 = 45 \, \text{ms}^{-1}$, $t_2 = 10 \, \text{s}$, and $a = 4 \, \text{ms}^{-2}$. $s_2 = u_2t_2 + \frac{1}{2}at_2^2$ $s_2 = (45)(10) + \frac{1}{2}(4)(10)^2$ $s_2 = 450 + 2(100)$ $s_2 = 450 + 200$ $s_2 = 650 \, \text{m}$

4. Identify the Answer

The distance traveled in the next 10 seconds is 650 m. This matches option A.

5. Exam Tip

When a question asks for the distance in the "next" time interval, always remember that the final velocity of the first interval becomes the initial velocity for the second interval. Don't make the mistake of using the original starting velocity for the second part of the journey!