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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 $a$.
[3]
$4 / 4$
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Answer from Sia

Posted about 1 month ago

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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