Question

3. A bike rider accelerates uniformly at 2.0 m/s2 for 10.0

s. If the rider starts from rest, calculate the distance
traveled in the fourth second.
(i.e. between t = 3 s and t = 4 s).

Answer 1

7 m

Step-by-step explanation:

The bike rider starts from rest. Initial velocity u = 0

Acceleration a = 2.0 m/s²

Total time taken = 10 s

Use the second equation of motion

s = ut + 0.5 at²

u = 0

s = 0.5at²

`s_4-s_3= 0.5a (t_4^2-t_3^2)`

Substitute the values

`s_4-s_3= 0.5* 2.0* (4^2-3^2) = 16-9 = 7 m`

Answer 2

Answer: 16 m

Step-by-step explanation:

Since this situation is about constanta acceleration, we can use the following equations:

`V=V_(o) +a.t` (1)

`V^(2)={V_(o)}^(2) +2ad` (2)

Where:

`V` is the bike rider's final velocity

`V_(o)=0` is the bike rider's initial velocity

`t=4 s` is the time at which we need to find the bike rider's distance

`a=2 (m)/(s^(2))` is the constant acceleration

`d` is the bike rider's traveled distance

From (1) we can find
`V`:

`V=0 +a.t` (3)

`V=(2 (m)/(s^(2)))(4 s)` (4)

`V=8(m)/(s)` (5)

Now, we can substitute (5) in (2) and find the traveled distance
`d`:

`V^(2)=0 +2ad` (6)

`d=(V^(2))/(2a)` (7)

Finally:

`d=16 m`