Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
Nov 2005 p4 q1
3361
A car travels in a straight line with constant acceleration \(a \text{ m s}^{-2}\). It passes the points \(A, B\) and \(C\), in this order, with speeds \(5 \text{ m s}^{-1}\), \(7 \text{ m s}^{-1}\) and \(8 \text{ m s}^{-1}\) respectively. The distances \(AB\) and \(BC\) are \(d_1 \text{ m}\) and \(d_2 \text{ m}\) respectively.
Write down an equation connecting
\(d_1\) and \(a\),
\(d_2\) and \(a\).
Hence find \(d_1\) in terms of \(d_2\).
Solution
To solve this problem, we use the equation of motion:
\(v^2 = u^2 + 2as\)
where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance.
(a) For the distance \(d_1\) between points \(A\) and \(B\):
Initial speed \(u = 5 \text{ m s}^{-1}\), final speed \(v = 7 \text{ m s}^{-1}\).
\(7^2 = 5^2 + 2ad_1\)
\(49 = 25 + 2ad_1\)
\(24 = 2ad_1\)
\(12 = ad_1\)
(b) For the distance \(d_2\) between points \(B\) and \(C\):
Initial speed \(u = 7 \text{ m s}^{-1}\), final speed \(v = 8 \text{ m s}^{-1}\).
