- #1

Warcus

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## 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