A bus moves at a constant speed of 11.1 m/s. Based solely on this information, can you determine if the bus is not accelerating or not? Explain your response in two lines.
Added by Ana S.
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1 m/s, this means that its velocity is not changing over time. Show more…
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You are walking toward the back of a bus that is moving forward with a constant velocity. Describe your motion relative to the bus and relative to a point on the ground. if I am walking toward the back of the bus at the same velocity it is traveling,is it correct to say that there is there no motion for me relative to the bus?
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Prabhu R.
A bus moving in a straight line at a speed of $20 \mathrm{~m} / \mathrm{s}$ begins to slow at a constant rate of $3.0 \mathrm{~m} / \mathrm{s}$ each second. Find how far it goes before stopping. Take the direction of motion to be the $+x$ -direction. For the trip under consideration, $u_{i}=20 \mathrm{~m} / \mathrm{s}, v_{f}=0 \mathrm{~m} / \mathrm{s}, a=-3.0 \mathrm{~m} / \mathrm{s}^{2}$. Notice that the bus is not speeding up in the positive motion direction. Instead, it is slowing in that direction and so its acceleration is negative (a deceleration). Use $$ v_{f x}^{2}=v_{i x}^{2}+2 a x \quad \text { and, hence, } \quad 0=(20 \mathrm{~m} / \mathrm{s})^{2}+2\left(-3.0 \mathrm{~m} / \mathrm{s}^{2}\right) x $$ to find $$ x=\frac{-(20 \mathrm{~m} / \mathrm{s})^{2}}{2\left(-3.0 \mathrm{~m} / \mathrm{s}^{2}\right)}=67 \mathrm{~m} $$
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