A particle moves along a straight line relative to a fixed origin $O$ with acceleration $a(t)=\sqrt{t^3}$, where $a(t)$ is measured in $m / s ^2$ and $t$ is measured in seconds. If the particle has velocity $v=0 m / s$ when $t=0 s$, calculate the speed of the particle at the moment $t=1 s$.
a $\frac{5}{2} m / s$
b $\frac{2}{5} m / s$
C $\frac{2}{3} m / s$
d $\frac{3}{2} m / s$
e $\frac{1}{5} m / s
Answer (1)
Integrate the acceleration function a ( t ) = t 2 3 to find the velocity function v ( t ) .
Apply the power rule for integration: v ( t ) = ∫ t 2 3 d t = 5 2 t 2 5 + C .
Use the initial condition v ( 0 ) = 0 to determine the constant of integration C = 0 .
Evaluate the velocity at t = 1 : v ( 1 ) = 5 2 ( 1 ) 2 5 = 5 2 m / s .
Explanation
Problem Setup We are given the acceleration function a ( t ) = t 3 = t 2 3 and the initial velocity v ( 0 ) = 0 . We need to find the speed of the particle at t = 1 .
Integrating Acceleration To find the velocity function v ( t ) , we integrate the acceleration function with respect to time: v ( t ) = ∫ a ( t ) d t = ∫ t 2 3 d t
Velocity Function Evaluating the integral, we get: v ( t ) = 2 5 t 2 5 + C = 5 2 t 2 5 + C
Finding the Constant of Integration Using the initial condition v ( 0 ) = 0 , we can find the constant of integration C :
0 = 5 2 ( 0 ) 2 5 + C ⟹ C = 0
Complete Velocity Function So the velocity function is: v ( t ) = 5 2 t 2 5
Calculating Speed at t=1 Now, we can find the speed at t = 1 by substituting t = 1 into the velocity function: v ( 1 ) = 5 2 ( 1 ) 2 5 = 5 2 s m
Final Answer The speed of the particle at t = 1 is 5 2 m / s .
Examples
Imagine you are designing a rocket launch sequence. Knowing the acceleration function of the rocket, you can determine its velocity at any given time. This is crucial for ensuring the rocket reaches the correct speed at different stages of the launch, allowing for precise trajectory adjustments and successful deployment of satellites or other payloads. By integrating the acceleration, you can predict the rocket's speed and position, optimizing the launch for maximum efficiency and accuracy.
