- Cyclist: Acceleration \(1.6 \text{ m/s}^2\), Distance \(35 \text{ m}\).
- Train: Deceleration \(1.0 \text{ m/s}^2\), Time \(30 \text{ s}\).
- Sprinter: Velocity at \(2.0 \text{ s}\) is \(8.0 \text{ m/s}\), Distance \(8.0 \text{ m}\), Total Distance \(56.0 \text{ m}\).
- Braking: Braking distance \(52.08 \text{ m}\), Total stopping distance \(70.83 \text{ m}\). Speed impacts distance quadratically.
- Vertical motion: Max height \(11.48 \text{ m}\), Time to max \(1.53 \text{ s}\), Total flight time \(4.61 \text{ s}\), Final velocity \(-30.18 \text{ m/s}\).
- Final Answers: \(\boxed{1.6 \text{ m/s}^2, 35 \text{ m}; 1.0 \text{ m/s}^2, 30 \text{ s}; 8.0 \text{ m/s}, 8.0 \text{ m}, 56.0 \text{ m}; 52.08 \text{ m}, 70.83 \text{ m}; 11.48 \text{ m}, 1.53 \text{ s}, -30.18 \text{ m/s}, 4.61 \text{ s}}\)
Explanation
- Problem Analysis and Kinematic Principles
The objective is to solve five kinematics problems involving uniform acceleration. We use standard kinematic equations:
- \(v = u + at\)
- \(s = ut + \frac{1}{2}at^2\)
- \(v^2 = u^2 + 2as\)
- \(s = \frac{u+v}{2}t\)
These equations relate displacement (\(s\)), initial velocity (\(u\)), final velocity (\(v\)), acceleration (\(a\)), and time (\(t\)). We treat upward or forward movement as positive and deceleration as negative acceleration.
- Solving Cyclist and Train Problems
Problem 3 (Cyclist):
- Given: \(u = 3.0 \text{ m/s}\), \(v = 11 \text{ m/s}\), \(t = 5.0 \text{ s}\).
- (a) Acceleration: \(a = \frac{v - u}{t} = \frac{11 - 3.0}{5.0} = \frac{8.0}{5.0} = 1.6 \text{ m/s}^2\).
- (b) Distance: \(s = \frac{u + v}{2}t = \frac{3.0 + 11}{2}(5.0) = 7.0 \times 5.0 = 35 \text{ m}\).
Problem 4 (Freight Train):
- Given: \(u = 30 \text{ m/s}\), \(v = 0 \text{ m/s}\), \(s = 450 \text{ m}\).
- (a) Deceleration: Using \(v^2 = u^2 + 2as\), \(0 = 30^2 + 2a(450)\), so \(900a = -900\), giving \(a = -1.0 \text{ m/s}^2\). The deceleration is \(1.0 \text{ m/s}^2\).
- (b) Time to stop: \(v = u + at \implies 0 = 30 + (-1.0)t\), so \(t = 30 \text{ s}\).
- Solving Sprinter Problem
Problem 5 (Sprinter):
- (a) Velocity at \(t = 2.0 \text{ s}\): \(v = u + at = 0 + (4.0)(2.0) = 8.0 \text{ m/s}\).
- (b) Distance during acceleration: \(s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(4.0)(2.0^2) = 2 \times 4 = 8.0 \text{ m}\).
- (c) Total distance at \(8.0 \text{ s}\): After \(2.0 \text{ s}\), the sprinter moves at a constant velocity of \(8.0 \text{ m/s}\) for the remaining \(6.0 \text{ s}\). Distance \(s_2 = v \times t_{rem} = 8.0 \times 6.0 = 48.0 \text{ m}\). Total distance \(s_{total} = s_1 + s_2 = 8.0 + 48.0 = 56.0 \text{ m}\).
- Solving Highway Braking Problem
Problem 6 (Highway Braking):
- (a) Braking distance (\(s_b\)): With \(u = 25 \text{ m/s}\), \(v = 0\), \(a = -6.0 \text{ m/s}^2\), \(v^2 = u^2 + 2as_b \implies 0 = 25^2 + 2(-6.0)s_b \implies 625 = 12s_b\). Thus, \(s_b \approx 52.08 \text{ m}\).
- (b) Total stopping distance: During reaction time (\(t_r = 0.75 \text{ s}\)), distance \(s_r = u \times t_r = 25 \times 0.75 = 18.75 \text{ m}\). Total distance \(S = s_r + s_b = 18.75 + 52.08 = 70.83 \text{ m}\).
- (c) Impact of speed: Stopping distance consists of reaction distance (\(\propto u\)) and braking distance (\(\propto u^2\)). Because braking distance increases quadratically with speed, even small increases in speed result in significant increases in the total stopping distance, making following distance critical.
- Solving Vertical Motion Problem
Problem 7 (Vertical Motion):
- (a) Max height above cliff: \(v=0, a=-9.8, u=15 \implies 0^2 = 15^2 + 2(-9.8)h \implies h = \frac{225}{19.6} \approx 11.48 \text{ m}\).
- (b) Time to max height: \(0 = 15 - 9.8t \implies t \approx 1.53 \text{ s}\).
- (c) Total flight time: Setting final position \(y = 0\) (sea level), \(y = y_0 + ut + \frac{1}{2}at^2 \implies 0 = 35 + 15t - 4.9t^2\). Solving via the quadratic formula gives \(t \approx 4.61 \text{ s}\).
- (d) Final velocity: \(v = u + at = 15 - 9.8(4.61) \approx -30.18 \text{ m/s}\).
Examples
Understanding these kinematics problems is vital for safety and engineering. For example, knowing the relationship between speed and stopping distance allows engineers to design roads with appropriate speed limits and signage. It also helps drivers recognize why maintaining a safe following distance, which grows quadratically with speed, is non-negotiable for preventing accidents.