Answer:
(a).The acceleration is 52.5 m/s²
(b). The acceleration is 5.35 g.
(c). The time is 0.038 sec.
Explanation:
Given that,
Initial speed = 2.0 m/s
Distance = 2.2 cm
Final speed = 1.3 m/s
We need to calculate the acceleration
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
Put the value into the formula
[tex]a=\dfrac{(1.3)^2-(2.0)^2}{2\times2.2\times10^{-2}}[/tex]
[tex]a=-52.5\ m/s^2[/tex]
Negative sign shows the deceleration
(b). We need to calculate the acceleration in terms g
[tex]a=\dfrac{52.5}{9.8}[/tex]
[tex]a=5.35\ g[/tex]
The acceleration is 5.35 g.
(c). We need to calculate the time
Using equation of motion
[tex]v=u+at[/tex]
[tex]t=\dfrac{v-u}{a}[/tex]
Put the value into the formula
[tex]t=\dfrac{0-2}{-52.5}[/tex]
[tex]t=0.038\ sec[/tex]
Hence, (a).The acceleration is 52.5 m/s²
(b). The acceleration is 5.35 g.
(c). The time is 0.038 sec.