Question

A car travelling at a speed of $30\mathrm{km}/\mathrm{h}$ is brought to a halt in 8 meters by applying brakes. If the same car is travelling at $60\mathrm{km}/\mathrm{h}$ it can be brought to a halt with the same breaking power in (A) $8\mathrm{m}$ (B) $16\mathrm{m}$ (C) $24\mathrm{m}$ (D) $32\mathrm{m}$

Short answer

The car traveling at 60 km/h can be brought to a halt with the same braking power in $32\mathrm{m}$.

Step-by-step solution

Convert speeds to m/s

To work with speeds in meters and seconds, we first need to convert the given speeds. Speed can be converted from km/h to m/s by multiplying it with 5/18: Speed at 30 km/h = $30\ast \frac{5}{18}$ m/s Speed at 60 km/h = $60\ast \frac{5}{18}$ m/s

Calculate deceleration

We can use the formula v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, deceleration), and s is the distance traveled. We know that v = 0 (since the car comes to a halt), u = 30 * 5/18 m/s, and s = 8 meters. Plugging these values into the formula, we can find the deceleration: $0=(30\ast \frac{5}{18}{)}^2+2a(8)$

Solve for deceleration

Now solve for "a": a = $-\frac{(30\ast \frac{5}{18}{)}^2}{2\ast 8}$

Use deceleration to find the stopping distance at 60 km/h

Now that we have the deceleration caused by braking, we can use the same formula (v^2 = u^2 + 2as) to find the stopping distance for the car traveling at 60 km/h: 0 = $(60\ast \frac{5}{18}{)}^2+2\cdot a\cdot s$ where "a" is the same deceleration found in step 3, and "s" is the distance we are looking for.

Solve for stopping distance

Rearrange the equation to solve for "s": s = $-\frac{(60\ast \frac{5}{18}{)}^2}{2a}$ By substituting the value of "a" from step 3, we can find the stopping distance "s": s = 32 m So the answer is (D) $32\mathrm{m}$.

Key concepts

Deceleration

In the realm of kinematics, deceleration refers to the process where an object slows down, which is essentially negative acceleration. When a car decelerates, it means the velocity is decreasing over time due to forces like braking. In the exercise given, the deceleration is calculated using the motion equation

• $$v^2=u^2+2as$$

Here, $v$ is the final velocity (0, since the car stops), $u$ is the initial velocity, $a$ is the deceleration, and $s$ is the distance. By rearranging the formula to solve for $a$, we find how quickly the car loses speed:

• a = $-\frac{u^2}{2s}$

This formula helps us understand how effective the brakes are in bringing the vehicle to a halt, allowing us to determine the influence of varying speeds on stopping performance.

Stopping Distance

The stopping distance of a vehicle refers to the distance it travels while coming to a stop after the brakes are applied. This is a crucial aspect in understanding vehicle safety and responsiveness. In the given exercise, the stopping distance at an initial speed of 30 km/h is 8 meters.

When the speed increases to 60 km/h, the kinetic energy and subsequently the stopping distance increase significantly. The motion equation

• $$s=-\frac{u^2}{2a}$$

is used to calculate the new stopping distance. Here, $s$ is the stopping distance, $u$ is the initial speed, and $a$ is the deceleration.

From this exercise, we see that doubling the speed results in a stopping distance of 32 meters, highlighting the quadratic relationship between speed and stopping distance - as speed doubles, stopping distance quadruples. Keeping this in mind is vital for both drivers and vehicle designers.

Motion Equations

Motion equations are essential tools in kinematics, allowing us to predict and analyze motion under various conditions. Among these equations,

• $$v^2=u^2+2as$$

proves particularly powerful for problems involving uniform acceleration or deceleration. Here, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $s$ is the distance covered.

This formula helps calculate how far or how fast an object will travel under a constant force. In the context of our exercise, it allows us to determine both the initial deceleration and the stopping distance under different conditions. Breaking down these equations into manageable parts can help better understand complex motions and apply them appropriately in real-world scenarios.

Using these equations empowers students and engineers to predict vehicle behavior and optimize design and road safety effectively.