a truck’s speed increases uniformly from 10m/s to 30m/s in 20 secs.
find the
a. average speed
b. acceleration
help me how will i be able to find the avg speed and acceleration? I am confused..help please..
to: curtis
in your solution to avg speed, what is
s0?
v0t?
at?
and where did u get 0.5at^2?
i'm sorry if im a bit asky, but it's my way to understand it..
can u kindly help me solve using this formula:
v (avg speed) = total distance / total time
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2 responses so far ↓
1 Curtis Strangelove III // Aug 3, 2008
I'll solve b) first.
Given v = v0 + at and v0 = 10 m/s:
30 m/s = 10 m/s + a (20s)
20 m/s = a (20s)
a = 1 m/s^2
Acceleration is 1 m/s^2.
a. s = s0 + v0t + 0.5at^2 (Let s0 = 0.)
s(20s) = (10 m/s)(20s) + 0.5(1m/s^2)(20s)^2
s = 200m + 200m = 400m
Average speed = 400m/20s = 20 m/s
ETA:
Empty, s0 is the initial position of the truck. That position is irrelevant to your question, so I set it at zero.
v0t is initial velocity (10m/s) multiplied by time.
at is acceleration multiplied by time.
"and where did u get 0.5at^2?"
It comes from calculus-based physics.
s = s0 + v0t + 0.5at^2 [ s is position ]
ds/dt = v = v0 + at [ v is velocity (Velocity is a vector quantity and speed is a scalar quantity.)]
dv/dt = a [ a is acceleration ]
"can u kindly help me solve using this formula:
v (avg speed) = total distance / total time"
I did. The total time was 20 seconds. The total distance was 400 meters, so the average speed is 20m/s.
2 ashleyjm // Aug 3, 2008
average speed=10m/s(i m not sure of dis)
nyways now for acceleration
a=v-u/t
where,
v=final velocity
u=initial velocity
t=time taken
a=30-10/20=20/20=1 second
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