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An object, initially at rest, moves with a constant acceleration of 10 m/s2. How far will it travel in (a) 2.0 s and (b) 4.0 s? If this object had an initial velocity of 4 m/s, how far will it travel in (C) 2.0 s and (d) 4.0 s?

User Fishingfon
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Answer:

(a) d = 20 m

(b) d' = 80 m

(c) x = 28 m

(d) x' = 96 m

Solution:

As per the question:

Initial velocity of the object, v = 0

Constant acceleration of the object,
a_(c) = 10 m/s^(2)

(a) Distance traveled, d in t = 2.0 s is given by the second eqn of motion:


d = vt + (1)/(2)a_(c)t^(2)


d = 0.t + (1)/(2)* 10* 2^(2) = 20 m

(b) Distance traveled, d' in t = 4.0 s is given by the second eqn of motion:


d' = vt + (1)/(2)a_(c)t^(2)


d' = 0.t + (1)/(2)* 10* 4^(2) = 80 m

Now, when initial velocity, v = 4 m/s, then

(c) Distance traveled, x in t = 2.0 s is given by the second eqn of motion:


x = vt + (1)/(2)a_(c)t^(2)


x = 4* 2.0 + (1)/(2)* 10* 2^(2) = 28 m

(d) Distance traveled, x' in t = 4.0 s is given by the second eqn of motion:


x = vt + (1)/(2)a_(c)t^(2)


x = 4* 4.0 + (1)/(2)* 10* 4^(2) = 96 m

User CybeX
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