Question

Greg and Tabitha are jogging together. At time t = 0, they are both traveling at a speed of 4 m/s, but Greg is 8 m ahead of Tabitha. Tabitha accelerates at a rate of 1.5 m/s2 and Greg accelerates at a rate of 1.0 m/s2.

a. Write an expression for Greg's position. Plug in all the values you can to simplify the expression. (1 point)

b. Write an expression for Tabitha's position. Plug in all the values you can to simplify the expression. (1 point)

c. How long does it take Tabitha to catch up to Greg? (1 point)

d. How far has Tabitha run while catching up to Greg? (1 point)

Answer 1

The answer to question is below

Step-by-step explanation:

Data

t1 = 0

vo = 4 m/s

a)

d = vot +
`(1)/(2) at^(2)`

d = 4t +
`(1)/(2) (1)t^(2) + 8`

`d = 8 + 4t + (t^(2) )/(2)`

b)

`d = 4t + (1)/(2) (1.5)t^(2)`

`d = 4t + 0.75t^(2)`

c)

`8 + 4t + (t^(2) )/(2)` =
`4t + 0.75t^(2)`

`4t + (1)/(2) t^(2) + 8 = 4t + (1)/(2) (1.5)t^(2) \\`

`0.75t^(2) - 0.5t^(2)  = 8`

`0.25t^(2) = 8`

t² = 32

t = 5.66 s

d)

d = 4(5.66) + 0.75(5.66)²

d = 22.64 + 24.02

d = 46.67 m