• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

projectile question (1 Viewer)

U MAD BRO

Member
Joined
Jul 29, 2012
Messages
287
Gender
Undisclosed
HSC
N/A
i have the distance, speed and time, how do i calculate the acceleration of a person running :$
the runner has to cover a distance of 50.36m in 2.86s, so i need to calculate his constant acceleration.
 

someth1ng

Retired Nov '14
Joined
Sep 18, 2010
Messages
5,558
Location
Adelaide, Australia
Gender
Male
HSC
2012
Uni Grad
2021
Use the formula: s=ut+(1/2)at^2

The formula sheet will have a different notation and because it's a lot harder to remember, I used that one.
 

U MAD BRO

Member
Joined
Jul 29, 2012
Messages
287
Gender
Undisclosed
HSC
N/A
Use the formula: s=ut+(1/2)at^2

The formula sheet will have a different notation and because it's a lot harder to remember, I used that one.
what would u sub s with?
and is u 0m/s in this case? since it is horizontal?
 

Rathaen

Member
Joined
Dec 26, 2011
Messages
55
Location
Sydney
Gender
Male
HSC
2011
s = displacement
u = initial velocity
a = acceleration
t = time
 

someth1ng

Retired Nov '14
Joined
Sep 18, 2010
Messages
5,558
Location
Adelaide, Australia
Gender
Male
HSC
2012
Uni Grad
2021
what would u sub s with?
and is u 0m/s in this case? since it is horizontal?
u=initial velocity, if they're starting rest rest then you sub in 0ms^-1
s=displacement, if they're starting from position 0m then the displacement at the end will be 50.36m
a and t are quite obvious.

Solution
s=50.36m
t=2.86s
u=0ms^-1

s=ut+(1/2)at^2
50.36=(0)(2.86)+(1/2)(2.86^2)a
a=[50.36]/[(1/2)(2.86^2)]
a=12.3ms^-2 in the initial direction
 
Last edited:

U MAD BRO

Member
Joined
Jul 29, 2012
Messages
287
Gender
Undisclosed
HSC
N/A
u=initial velocity, if they're starting rest rest then you sub in 0ms^-1
s=displacement, if they're starting from position 0m then the displacement at the end will be 50.36m
a and t are quite obvious.
thanks =)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top