A car traveling due east with an initial velocity of 10 m/s accelerates for 8 s

Question

A car traveling due east with an initial velocity of 10 m/s accelerates for 8 s for a distance of 150 m. What is the velocity of the car after the acceleration? Sue walks down the road from her home with a velocity of 8.5 cm/s and having a constant acceleration of 0.6 cm/s^2. How long did it take her to traverse a distance of 2 m? Consider a particle who's position is determined by the equation x(t) = 10 m + (2 m/s) t + (1.12 m/s^2) t^2 - (0.1 m/s^3) t^3. (a) What are the expressions for the particle's velocity and acceleration as a function of time? (b) At what time is the velocity zero? (c) At what time is the velocity a maximum?

Explanation / Answer

1)
vi = 10 m/s
t = 8 s
d = 150 m
1st find the acceleration using:
d = vi*t + 0.5*a*t^2
150 = 10*8 + 0.5*a*8^2
150 = 80 + 32*a
a = 2.1875 m/s^2

now calculate the final velocity using:
vf = vi + a*t
= 10 + 2.1875*8
= 27.5 m/s
Answer: 27.5 m/s

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