- #1
Warcus
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Hi there everyone! I'm a first-year physics student and I need some help with a couple of questions! OK, here it goes :)
the car has a speed of 5.0 m/s, what is the velocity at t=3.0 s?
(α = alpha) (In other words a(t)=α=1.4 m/s^3 is not a typo)
There are four possible answers:
(a) y (3) = 6.3 m / s
(b) v (3) = 7.4 m / s
(c) y (3) = 8.7 m / s
(d) y (3) = 11 m / s
position at t = 1,0s is 5.0 m Find the car's position at t = 3.0 s
There are four possible answers:
(a) x (3) = 15m
(b) x (3) = 18m
(c) x (3) = 20m
(d) x (3) = 23m
The relevant equations would be the kinematic equations for constant acceleration:
V = Vo + at
X = Xo + VoT+ 0.5aT^2
V^2=Vo^2 + 2a(-Xo)
V = 5 m/s + 1,4 m/s^2 *2 s (the difference between the two t-values is 2)
V = 7,8 m/s
This answer is not one of the possible solutions --> I have done something wrong
Possible solutions (again):
(a) y (3) = 6.3 m / s
(b) v (3) = 7.4 m / s
(c) y (3) = 8.7 m / s
(d) y (3) = 11 m / s
X=VoT + 0,5at^2
X=5 m/s x 2 s + 0,5*1,4 m/s^2 * (2s)^2 (Again, the difference between the t-values = 2)
X = 12,8 m
The task specifies that the distance at T(1)=5 --> Total distance = 5 m + 12,8 m = 17,8
Again, ther are four possible answers, and I'm fairly sure that the answer is b:
(a) x (3) = 15m
(b) x (3) = 18m
(c) x (3) = 20m
(d) x (3) = 23m
Nevertheless, I am a bit unsure about my calculation, and posted this Task as well :)
Thanks a lot for your help! :D
Homework Statement
Task 1
The acceleration of a vehicle is given by a(t) = αT, where α=1.4 m/s^3. At the time t=1.0 sthe car has a speed of 5.0 m/s, what is the velocity at t=3.0 s?
(α = alpha) (In other words a(t)=α=1.4 m/s^3 is not a typo)
There are four possible answers:
(a) y (3) = 6.3 m / s
(b) v (3) = 7.4 m / s
(c) y (3) = 8.7 m / s
(d) y (3) = 11 m / s
Task 2
We use with the same car as in the previous problem. We learn that the carposition at t = 1,0s is 5.0 m Find the car's position at t = 3.0 s
There are four possible answers:
(a) x (3) = 15m
(b) x (3) = 18m
(c) x (3) = 20m
(d) x (3) = 23m
Homework Equations
The relevant equations would be the kinematic equations for constant acceleration:
V = Vo + at
X = Xo + VoT+ 0.5aT^2
V^2=Vo^2 + 2a(-Xo)
The Attempt at a Solution
Task 1
V = Vo + atV = 5 m/s + 1,4 m/s^2 *2 s (the difference between the two t-values is 2)
V = 7,8 m/s
This answer is not one of the possible solutions --> I have done something wrong
Possible solutions (again):
(a) y (3) = 6.3 m / s
(b) v (3) = 7.4 m / s
(c) y (3) = 8.7 m / s
(d) y (3) = 11 m / s
Task 2
X=VoT + 0,5at^2
X=5 m/s x 2 s + 0,5*1,4 m/s^2 * (2s)^2 (Again, the difference between the t-values = 2)
X = 12,8 m
The task specifies that the distance at T(1)=5 --> Total distance = 5 m + 12,8 m = 17,8
Again, ther are four possible answers, and I'm fairly sure that the answer is b:
(a) x (3) = 15m
(b) x (3) = 18m
(c) x (3) = 20m
(d) x (3) = 23m
Nevertheless, I am a bit unsure about my calculation, and posted this Task as well :)
Thanks a lot for your help! :D