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xiaoxing's Question
Physics M
Posted about 1 month ago

4 A car starts from rest and moves in a straight line with constant acceleration a m s2a \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 2a m s22 a \mathrm{~m} \mathrm{~s}^{-2}.
(a) Sketch the velocity-time graph for the motion of the car.
[1]
(b) Find the value of aa.
[3]
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Answer from Sia
Posted about 1 month ago
Solution
a
The car starts from rest, so the initial velocity vi=0 v_i = 0 . It accelerates with a constant acceleration a a for a distance of 50 m. Using the equation of motion d=vit+12at2 d = v_i t + \frac{1}{2} a t^2 , we can express the time taken to cover this distance as: 50=0t+12at2    50=12at2    at2=100    t2=100a    t=100a50 = 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 vf=vi+at v_f = v_i + at : vf=0+a100a=10av_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 vc v_c can be calculated as: vc=500 m25 s=20 m/sv_c = \frac{500 \text{ m}}{25 \text{ s}} = 20 \text{ m/s}. Setting this equal to the final velocity from the previous phase gives: 10a=20    a=2    a=4 m/s210\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 m/s2 2a = 8 \text{ m/s}^2 . The time taken to decelerate from vc=20 m/s v_c = 20 \text{ m/s} to rest can be calculated using vf=vi+at v_f = v_i + at : 0=208t    8t=20    t=208=2.5 s0 = 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 a . The velocity-time graph reflects these phases accurately.

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