Question

1. in an incident on the m1 motorway in england, a jaguar sports car made a skid 276 m long while braking to a stop. assuming a constant deceleration of -8 m/s², what was the car’s initial speed?

Explanation:

Step1: Identify the kinematic equation

We use the kinematic equation \(v^2 = u^2 + 2as\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the displacement. Here, the car comes to a stop, so \(v = 0\space m/s\), \(a=- 8\space m/s^{2}\), and \(s = 276\space m\). We need to solve for \(u\).

Step2: Rearrange the equation to solve for \(u\)

Rearranging \(v^2=u^2 + 2as\) for \(u\), we get \(u^2=v^2 - 2as\). Substituting the known values: \(v = 0\), \(a=-8\), \(s = 276\). So \(u^2=0-2\times(- 8)\times276\).

Step3: Calculate the value of \(u^2\)

First, calculate \(2\times(-8)\times276=- 4416\), then \(u^2=0 - (-4416)=4416\).

Step4: Find the value of \(u\)

Take the square root of \(4416\) to find \(u\). \(u=\sqrt{4416}\approx66.45\space m/s\) (we take the positive root because speed is a scalar quantity and we are interested in the magnitude of the initial velocity).

Answer:

The initial speed of the car is approximately \(\boldsymbol{66.45\space m/s}\) (or we can also convert it to \(km/h\) by multiplying by \(\frac{3600}{1000}\), \(66.45\times3.6\approx239.22\space km/h\) if needed, but based on the calculation from the kinematic equation, the initial speed in \(m/s\) is approximately \(66.45\space m/s\)).