The magnitude of his accelerations of the bush baby is about 140 m/s² or 14g
Further explanation
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration ( m/s² )
v = final velocity ( m/s )
u = initial velocity ( m/s )
t = time taken ( s )
d = distance ( m )
Let us now tackle the problem!
Given:
h = 2.3 m
d = 0.16 m
Unknown:
a = ?
Solution:
At first we can calculate the takeoff speed of the bush baby.
[tex]v^2 = u^2 - 2gh[/tex]
[tex]0^2 = u^2 - 2 \times 9.8 \times 2.3[/tex]
[tex]u^2 = 45.08[/tex]
Next, we can calculate the acceleration in order to reach this takeoff speed.
[tex]v^2 = u^2 + 2ad[/tex]
[tex]45.08 = 0^2 + 2 \times a \times 0.16[/tex]
[tex]a = 45.08 \div 0.32[/tex]
[tex]a = 140.875 ~ m/s^2 \approx 140 ~ m/s^2[/tex]
[tex]a = 140.875 / 9.8 ~g = 14.375g \approx 14g[/tex]
Learn more
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
Answer details
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle
