JEE MAIN - Physics (2024 - 4th April Evening Shift - No. 23)
Explanation
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.
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