Question

A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 80 m at t2 = 5.0 s . Answer the following by considering ratios, without computing the car's velocity. Part A What is the car's position at

t = 2.5 s ?
Part B What will be its position at
t = 15 s ?

Answer 1

The car's position at t = 2.5 s will be 40 m as it is half of the interval to reach 80 m. At t = 15 s, three times the interval, the car will be at 240 m assuming constant velocity.

Step-by-step explanation:

Given that the car moves from x1 = 0 m at t1 = 0 s to x2 = 80 m at t2 = 5.0 s, we can determine its position at other times under the assumption of constant velocity. Because the velocity is constant, the ratios of times to positions are constant as well.

Part A: Car's Position at t = 2.5 s

The time t = 2.5 s is exactly half of the time interval given (5.0 s), so the position should be half of 80 m, which is 40 m.

Part B: Car's Position at t = 15 s

To find the position at t = 15 s, let's first determine the time ratio. The ratio of 15 s to 5.0 s is 3:1. Since the velocity is constant, the positions scale with time. So, the position x at t = 15 s is three times 80 m, giving us 240 m.

Answer 2

Part A) x=40m at t= 2.5s

Part B) x=240m at t=15 s

Step-by-step explanation:

Data:

x₁=0 , t₁=0

x₂=80m, t₂=5 s

t₃= 2.5 s

t₄= 15 s

problem development

We set the ratios x / t:

Part A)

`(x_(3) )/(t_(3) ) =(x_(2) )/(t_(2) )`

`(x_(3) )/(2.5) =(80)/(5 )`

x₃=16*2.5

x₃=40m

Part B)

`(x_(4) )/(t_(4) ) =(80 )/(5 )`

`(x_(4) )/(15) =(80)/(5 )`

x₄=16*15

x₄=240m