Question and answer
A block slides along a frictionless surface with a constant acceleration of 2 m/s2. At time t = 0 s the block is at x = 5m and travelling with a velocity of 3 m/s. a) Where is the block at t = 2 seconds? b) What is the block’s velocity at 2 seconds? c) Where is the block when it’s velocity is 10 m/s? d) How long did it take to get to this point?
Here is an illustration of the setup. The variables we know are: x0 = 5 m v0 = 3 m/s a = 2 m/s2 Part a) x = x0 + v0t + ½at2 Substitute t = 2 seconds for t and the appropriate values of x0 and v0. x = 5 m + (3 m/s)(2 s) + ½(2 m/s2)(2 s)2 x = 5 m + 6 m + 4 m x = 15 m The block is at the 15 meter mark at t = 2 seconds.
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Question|Asked by Taurus28
Asked 14 days ago|2/9/2024 12:09:33 PM
Updated 14 days ago|2/9/2024 12:32:06 PM
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Here is an illustration of the setup.

The variables we know are:
x0 = 5 m
v0 = 3 m/s
a = 2 m/s2

Part a)
x = x0 + v0t + ½at2

Substitute t = 2 seconds for t and the appropriate values of x0 and v0.

x = 5 m + (3 m/s)(2 s) + ½(2 m/s2)(2 s)2
x = 5 m + 6 m + 4 m
x = 15 m

The block is at the 15 meter mark at t = 2 seconds.
Added 14 days ago|2/9/2024 12:29:57 PM
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Part b)
v = v0 + at
v = (3 m/s) + (2 m/s2)(2 s)
v = 3 m/s + 4 m/s
v = 7 m/s

The block is travelling 7 m/s at t = 2 seconds.
Added 14 days ago|2/9/2024 12:30:19 PM
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Part C

v^2 = v0^2 + 2a(x – x0)
(10 m/s)^2 = (3 m/s)^2 + 2(2 m/s2)(x – 5 m)
100 m^2/s^2 = 9 m^2/s^2 + 4 m/s^2(x – 5 m)
91 m^2/s^2 = 4 m/s^2(x – 5 m)
22.75 m = x – 5 m
27.75 m = x

The block is at the 27.75 m mark.
Added 14 days ago|2/9/2024 12:31:42 PM
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Part D
v = v0 + at
10 m/s = 3 m/s + (2 m/s2)t
7 m/s = (2 m/s2)t
7 2 s = t

It takes 7 2 s or 3.5 s to get to the 27.75 m mark.
Added 14 days ago|2/9/2024 12:32:06 PM
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