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

Answer: 16 m

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

$a=2 \frac{m}{s^{2}}$ is the constant acceleration

$d$ is the bike rider's traveled distance

From (1) we can find $V$:

 $V=0 +a.t$  (3)

 $V=(2 \frac{m}{s^{2}})(4 s)$  (4)

 $V=8\frac{m}{s}$  (5)

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

$V^{2}=0 +2ad$ (6)

$d=\frac{V^{2}}{2a}$ (7)

Finally:

$d=16 m$

Answer 2

Answer:

7 m

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

$s_4-s_3= 0.5\times 2.0\times (4^2-3^2) = 16-9 = 7 m$