Answer:
a_x(3.98) = 7.0844 m/s^2
Explanation:
Given:
- The eastward component of car's velocity is given as follows:
v_x (t) = 0.890*t^2
For, 0 s < t < 5 s
Find:
What is the acceleration of the car when v_x = 14.1 m/s ?
Solution:
- We will first compute the time t at which the velocity v_x component equals 14.1 m /s :
14.1 = 0.890*t^2
t = sqrt (15.8427)
t = 3.98 s
- Now use the relation of v_x and derivative to obtain a_x acceleration in eastward direction:
a_x (t) = dv_x / dt
a_x(t) = 1.78*t
- Evaluate at t = 3.98 s:
a_x(3.98) = 1.783.98
a_x(3.98) = 7.0844 m/s^2
Explanation:
The motion of a particle is defined by the relation as:
........(1)
Differentiating equation (1) wrt t we get:
............(2)
When v = 2 ft/s
t₁ = 2.26 s
and t₂ = 0.73 s
Put the value of t₁ in equation (1) as :
x₁ = 14.23 ft
Put the value of t₂ in equation (1) as :
x₁ = 14.74 ft
For acceleration differentiate equation (2) wrt t as :
a = 12 t - 18.........(3)
Put t₁ and t₂ in equation 3 one by one as :
Hence, this is the required solution.