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A student walks from her apartment a distance d=139m to the bus stop at a speed of v1= 1.40m/s.
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A student walks from her apartment a distance d=139m to the bus stop at a speed of v1= 1.40m/s. The she realizes that she has
Top answer:
Her DISTANCE was 417 BUT Her DISPLACEMENT was 139 going to the stop and back is a displacement of
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A student is running at her top speed of 4.6 m/s to catch a bus, which is stopped at the bus stop. When the student is still a
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A bus on a straight road starts from rest at a bus stop and accelerates at 2 m/s2 until it
reaches a speed of 40 m/s. Then the
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in general v = Vi + a t x = Xi + Vi t + (1/2) a t^2 Phase 1, stop 1 to stop 2 v= 0 + 2 t = 40 x = 0
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-b+c*t
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A man travels 3/8 of his journey by train 3/5 of it by bus and walks the rest of the distance.If he walks 2 km. How much
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rest of distance = 1 - (3/8 + 3/5) = 1 - 39/40 = 1/40 (1/40)x = 2 x = 80 km so he went (3/5)(80) km
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vmax is just after the 11seconds vmax= a*t distance covered: break it up into three times
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When she wants to go, downtown, a woman walks 1/6 mile to a bus stop and catches a city bus. Last month, she walked a total of
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Let's assume the woman caught the city bus x times last month. If she walks 1/6 mile to the bus stop
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Jack is 12m from the bus and is running towards the bus at a constant 4.0m/s. If the bus accelerates away from the bus stop at
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To determine if Jack can catch the bus, let's calculate the time it takes for the bus to move away
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Jack is 12m from the bus and is running towards the bus at a constant 4.0m/s. If the bus accelerates away from the bus stop at
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bus goes distance d Jack goes distance (d+12) d = (1/2) a t^2 = .4 t^2 d+12 = 4 t so d = 4 t - 12 4
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A bus accelerates at 2.4 m/s2 from rest for 10 s. It then travels at constant speed for 25 s, after which it slows to a stop
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V1 = a*T1 = 2.4 * 10 = 24 m/s. d1 = 0.5a*T1^2 = 1.2*10^2 = 120 m. a. d2 = V1*T2 = 24 * 25 = 600 m.
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