A bus is begins to move with an acceleration of 1 m/s2. A boy who is 48 m behind the bus starts running at 10 m/s towards the bus. After what time bus will cross the boy?

To determine when the bus will cross the boy, we need to analyze the motion of both the bus and the boy. The bus starts from rest and accelerates, while the boy runs towards the bus at a constant speed. Let's break this down step by step.

Understanding the Motion of the Bus

The bus accelerates from rest with an acceleration of 1 m/s². We can use the equation of motion to find the distance it travels over time:

• Distance of the bus (d_bus): The formula for distance under constant acceleration is:

d_bus = ut + (1/2)at²

Here, u (initial velocity) is 0 (since the bus starts from rest), a is 1 m/s², and t is the time in seconds. Thus, the distance covered by the bus becomes:

d_bus = 0 + (1/2)(1)t² = 0.5t²

Analyzing the Boy's Motion

The boy starts 48 meters behind the bus and runs towards it at a constant speed of 10 m/s. The distance he covers over time can be expressed as:

• Distance of the boy (d_boy): The formula for distance at constant speed is:

d_boy = speed × time = 10t

Setting Up the Equation

To find out when the bus crosses the boy, we need to set the distances equal to each other, taking into account the initial distance of 48 meters between them:

Distance of the bus = Distance of the boy + 48

Substituting the equations we derived:

0.5t² = 10t + 48

Rearranging the Equation

Now, let's rearrange this equation to form a standard quadratic equation:

0.5t² - 10t - 48 = 0

To eliminate the decimal, we can multiply the entire equation by 2:

t² - 20t - 96 = 0

Solving the Quadratic Equation

We can solve this quadratic equation using the quadratic formula:

t = [ -b ± √(b² - 4ac) ] / 2a

In our case, a = 1, b = -20, and c = -96:

• Calculate the discriminant: b² - 4ac = (-20)² - 4(1)(-96) = 400 + 384 = 784
• Now, plug the values into the quadratic formula:

t = [20 ± √784] / 2

Since √784 = 28, we have:

t = [20 ± 28] / 2

This gives us two potential solutions:

• t = (20 + 28) / 2 = 24 seconds
• t = (20 - 28) / 2 = -4 seconds (not physically meaningful)

Final Result

The only valid solution is t = 24 seconds. Therefore, the bus will cross the boy after 24 seconds of motion.