4 A car starts from rest and moves in a straight line with constant acceleration $a \mathrm{~m} \mathrm{~s}^{-2}$ for a distance of 50 m . The car then travels with constant velocity for 500 m for a period of 25 s , before decelerating to rest. The magnitude of this deceleration is $2 a \mathrm{~m} \mathrm{~s}^{-2}$.
(a) Sketch the velocity-time graph for the motion of the car.
[1]
(b) Find the value of $a$.
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$4 / 4$

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Accepted Answer

∻Solution∻ ‖ a ⋮ The car starts from rest, so the initial velocity $ v_i = 0 $. It accelerates with a constant acceleration $ a $ for a distance of 50 m. Using the equation of motion $ d = v_i t + \frac{1}{2} a t^2 $, we can express the time taken to cover this distance as: $$50 = 0 \cdot t + \frac{1}{2} a t^2 \implies 50 = \frac{1}{2} a t^2 \implies a t^2 = 100 \implies t^2 = \frac{100}{a} \implies t = \sqrt{\frac{100}{a}}$$. The final velocity after this acceleration phase can be found using $ v_f = v_i + at $: $$v_f = 0 + a \cdot \sqrt{\frac{100}{a}} = 10\sqrt{a}$$. ‖ ‖ b ⋮ After accelerating, the car travels with constant velocity for 500 m over a period of 25 s. The constant velocity $ v_c $ can be calculated as: $$v_c = \frac{500 \text{ m}}{25 \text{ s}} = 20 \text{ m/s}$$. Setting this equal to the final velocity from the previous phase gives: $$10\sqrt{a} = 20 \implies \sqrt{a} = 2 \implies a = 4 \text{ m/s}^2$$. ‖ ‖ c ⋮ The car then decelerates to rest with a deceleration of $ 2a = 8 \text{ m/s}^2 $. The time taken to decelerate from $ v_c = 20 \text{ m/s} $ to rest can be calculated using $ v_f = v_i + at $: $$0 = 20 - 8t \implies 8t = 20 \implies t = \frac{20}{8} = 2.5 \text{ s}$$. The entire motion can now be represented in the velocity-time graph as three segments: acceleration, constant velocity, and deceleration. ‖ ∻Answer∻ ⚹ 4 m/s² ⚹ ∻Key Concept∻ ⚹ Kinematics: The study of motion without considering the forces that cause it. Key equations include the equations of motion for constant acceleration. ⚹ ∻Explanation∻ ⚹ The solution involves calculating the acceleration, constant velocity, and deceleration phases of the car's motion, leading to the final value of $ a $. The velocity-time graph reflects these phases accurately. ⚹

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