Kinematics and Dynamics of Car Braking and Free Falling Object
<p>1) To find the total distance the car travels during braking and the time it takes:</p>
<p>a) The deceleration \( a \) can be calculated using the relation \( a = \frac{{v^2 - u^2}}{{2s}} \), where initial velocity \( u = 50 \) m/s (since the car is initially moving at 50 km/h, we convert this to m/s by multiplying by \( \frac{{1000}}{{3600}} \)), final velocity \( v = 0 \) m/s (since the car stops), and \( s \) is the distance. But since we are given a 10% negative gradient, the effective deceleration is the sum of the deceleration due to braking and the deceleration due to the slope: \( a = a_{braking} + a_{slope} \).</p>
<p>b) The deceleration due to the slope \( a_{slope} \) is given by \( a_{slope} = g\sin(\theta) \), where \( g = 9.81 \) m/s\(^2\) is the acceleration due to gravity, and \( \sin(\theta) \) can be approximated by the slope percentage over 100, thus \( \sin(\theta) \approx \frac{{10}}{{100}} = 0.1 \).</p>
<p>c) Hence, \( a_{slope} = 9.81 \times 0.1 = 0.981 \) m/s\(^2\), and using \( a_{braking} = -0.3g = -0.3 \times 9.81 \) m/s\(^2\), we find \( a = a_{braking} + a_{slope} = -2.943 + 0.981 = -1.962 \) m/s\(^2\).</p>
<p>d) Now, we can calculate the distance \( s \) using \( s = \frac{{v^2 - u^2}}{{2a}} = \frac{{0^2 - (50/3.6)^2}}{{2 \times (-1.962)}} \).</p>
<p>e) The time \( t \) it takes to stop is given by \( t = \frac{{v - u}}{{a}} = \frac{{0 - (50/3.6)}}{{-1.962}} \).</p>
<p>2) To find the speed at which the body impacts the water:</p>
<p>a) Use the kinetic energy at impact \( K.E. = \frac{1}{2}mv^2 \) equating it to the potential energy at the start \( P.E. = mgh \).</p>
<p>b) Since \( K.E. = P.E. \), we have \( \frac{1}{2}mv^2 = mgh \). After canceling mass \( m \), the equation simplifies to \( v^2 = 2gh \).</p>
<p>c) Plug in \( g = 9.81 \) m/s\(^2\) and \( h = 10 \) m to find \( v = \sqrt{2 \times 9.81 \times 10} \).</p>
<p>The calculations from the above steps will give you the distance the car travels during braking and the time it takes to stop, as well as the speed at which the object hits the water.</p>
