Question

An object's motion can be described by the equation x(t) = (6.0 m/s²)t²-(2.0 m/s)t+1.0 m. 

a. Find the object's position, velocity, and acceleration at t = 3.0 s. Plot the object's position, velocity, and acceleration (all on separate plots) from t = 0 s to t= 5 s. 

The following graph shows an object's velocity over time. At t-0 s the objects begins its motion at x = -2m.

b. What is the object's displacement from t=0 s to t = 6 s? 

c. what is the object's position at t = 6 s? 

d. What is the object's acceleration at t = 1 s,t = 2s, t = 4 s, t = 5.5s?

Answer 1

An object's motion can be described by the equation which given below as -

x (t) = (6 m/s²) t² - (2 m/s) t + (1 m)                                                            { eq.1 }

a. At t = 3 sec, the object's position which will be given as -

x = [(6 m/s²) (3 s)² - (2 m/s) (3 s) + (1 m)]

x = [(54 m) - (6 m) + (1 m)]

x = 49 m

Differentiating eq.1 w.r.t time, then we get

d [x (t)] / dt = 2 (6 m/s²) t - (2 m/s)                

v (t) = (12 m/s²) t - (2 m/s)                                                                                   { eq.2 }

At t = 3 sec, the object's velocity which will be given as -

v = [(12 m/s²) (3 s) - (2 m/s)]

v = [(36 m/s) - (2 m/s)]

v = 34 m/s

Differentiating eq.2 w.r.t time, then we get

d [v (t)] / dt = (12 m/s²)

a = (12 m/s²)

b. The object's displacement from t=0 sec to t=6 sec which will be given as -

d = d₀₁ + d₁₃ + d₃₅ + d₅₆

d = [(1/2) (1 s) (1 m/s)] + [(1/2) (2 s) (2 m/s)] + [(2 s) (2 m/s)] + [(1/2) (1 s) (2 m/s)]

d = [(0.5 m) + (2 m) + (4 m) + (1 m)]

d = 7.5 m

c. The object's position which at t = 6 sec which will be given as -

x (t) = (6 m/s²) t² - (2 m/s) t + (1 m)

x = [(6 m/s²) (6 s)² - (2 m/s) (6 s) + (1 m)]

x = [(216 m) - (12 m) + + (1 m)]

x = 205 m