Answer :
Answer:
a) 240 m
b) 122 m/s
c) 28 m/s²
Explanation:
Given:
Equation for motion
x = 38t + 14t²
a) average velocity during first 3 seconds
average velocity = [tex]\frac{\textup{change in displacement}}{\textup{cahnge in time}}[/tex]
now,
distance, at t = 0 s
x = 38 × 0 + 14 × 0² = 0 m
distance, at t = 3 s
x = 38 × 3 + 14 × 3² = 240 m
therefore,
average velocity = [tex]\frac{240-0}{3-0}[/tex] = 80 m/s
b) instantaneous velocity of the proton at t = 3.0 s
Instantaneous velocity, v = [tex]\frac{dx}{dt}=38+28\times t[/tex]
or
Instantaneous velocity, v = [tex]\frac{dx}{dt}=38+28\times 3[/tex]
= 122 m/s
c) instantaneous acceleration of the proton at t = 3.0 s
Now,
Acceleration = [tex]\frac{dv}{dt}[/tex] = 0 + 28 = 28 m/s²