Question 2: A car starts from rest and travels at a constant acceleration of 3000mi/h^2 along a straight path.How long does it take for it to reach a distance of 1 mile? What is its velocity at that point? (a) 1.55 min, 60mi/h (b) 3.33min,77.5mi/h (c) 3.33 min, 60mi/h (d) 1.55min,77.5mi/h

To solve this problem, we can use the kinematic equations for uniformly accelerated motion. The car starts from rest, so the initial velocity \( u = 0 \). The acceleration is given as \( a = 3000 \, \text{mi/h}^2 \). We need to find the time \( t \) it takes to travel a distance \( s = 1 \) mile and the velocity \( v \) at that distance.<br /><br />First, let's find the time \( t \) using the equation:<br />\[ s = ut + \frac{1}{2}at^2 \]<br /><br />Since \( u = 0 \), the equation simplifies to:<br />\[ s = \frac{1}{2}at^2 \]<br /><br />Solving for \( t \):<br />\[ t^2 = \frac{2s}{a} \]<br />\[ t^2 = \frac{2 \times 1 \, \text{mile}}{3000 \, \text{mi/h}^2} \]<br />\[ t^2 = \frac{2}{3000} \]<br />\[ t^2 = \frac{1}{1500} \]<br />\[ t = \sqrt{\frac{1}{1500}} \]<br />\[ t \approx \frac{1}{38.73} \]<br />\[ t \approx 0.0258 \, \text{hours} \]<br />\[ t \approx 1.55 \, \text{minutes} \]<br /><br />Next, let's find the velocity \( v \) at that distance using the equation:<br />\[ v = u + at \]<br /><br />Since \( u = 0 \), the equation simplifies to:<br />\[ v = at \]<br /><br />Substituting the values:<br />\[ v = 3000 \, \text{mi/h}^2 \times 0.0258 \, \text{hours} \]<br />\[ v \approx 77.5 \, \text{mi/h} \]<br /><br />Therefore, the correct answer is:<br />(d) $1.55min,77.5mi/h$