QJEEM24Phe192: JEE Main 2024 Physics - Class 11 JEEM24Phe192 Solution: A bus moving along a straight highway with speed of is brought…

A bus traveling along a straight highway at a speed of 72 km/h comes to a stop within 4 seconds after the brakes are applied. To find the distance the bus travels during this time (assuming uniform deceleration), follow these steps:

First, convert the initial speed from km/h to m/s:

$\begin{aligned} u &= 72 \times \frac{5}{18} = 20 \, \text{m/s} \end{aligned}$

Given:

Initial speed ( $u$ ) = 20 m/s

Final speed ( $v$ ) = 0 m/s

Time ( $t$ ) = 4 s

Using the first equation of motion:

$\begin{aligned} v &= u + at \\ 0 &= 20 + 4a \\ a &= -5 \, \text{m/s}^2 \end{aligned}$

Now, use the second equation of motion to find the distance traveled ( $S$ ):

$\begin{aligned} S &= ut + \frac{1}{2}at^2 \\ S &= 20 \times 4 + \frac{1}{2} \times (-5) \times 4^2 \\ S &= 80 - 40 \\ S &= 40 \, \text{m} \end{aligned}$

Therefore, the distance traveled by the bus during the time the brakes are applied is 40 meters.