Question

Evaluate the velocity of a body and its distance moved, both after 2 minutes from it starts from rest, when its acceleration is given by a = 15 m/s^2

Answer 1

Clicked closed by accident on last post :/

Answer 2

Any help? :( this if for calculus study

Answer 3

acceleration = dv/dt = 15 constant
separating variables
dv = 15dt

integrating v = 15t + c

when t = 0 v = 0 ( as it starts from rest) so c = 0

v = 15 t

so substitute t = 120 to find the velocity after 2 minutes

Answer 4

ok?

Answer 5

So v = 120 x 15 =1800 m/s?

Answer 6

yes

Answer 7

Now how will I go about finding the distance moved?

Answer 8

Ps thanks for the help - i struggle with calculus haha

Answer 9

a further integration will give you the distance travelled in terms of t
the constant of integration will again be 0 because it was a start from rest

Answer 10

cwrw238 asked me to look at this for you as she had to go to sit an examination

Answer 11

integrating will give you distance = 15 * t^2/2

Answer 12

So is the answer 108000m^2 or 24300000m^2 :/

Answer 13

Oh that was nice of her and nice of you to help :)

Answer 14

she made a mistake
she put t = 1800 instead of t = 120

answer is 1/2 * 15 * 120^2
= 108,000 meters

Answer 15

i'll kid her about that when i see her next!!
its not like her to make mistakes - but we all do!

Answer 16

Ahh I see that's what was getting me soo confused lol. That's true! Haha well thank you both for your help!! I'm so happy to have got that question done :) :)

Answer 17

yw