John was traveling in a car at 30 m/s in the direction east. Starting at t=0s he

Question

John was traveling in a car at 30 m/s in the direction east. Starting at t=0s he slowed his car down to 5 m/s within 35s. After slowing the car down, he made a turn to the left, and another 30s later was traveling due north. During the turn, between t=35s and t=65s, the magnitude of john's acceleration was 0.5 m/s^2.

What was the magnitude of John's acceleration in as he was slowing down?

What was the x-component ("east"-component) of John's acceleration during the turn?

What was the y-component of John's acceleration during the turn?

After completeing the turn, how fast was John moving?

Explanation / Answer

1)

acceleration , a = v-u/t

= (5-30)/35

= 25/35

= 0.714 m/s2

2)

Let x-component of acceleration during turn be ax

Now,

a = sqrt(ax^2+ay^2) = 0.5

ax = 5/t = 5/30

ay = v/t

So, sqrt((5/30)^2 + (v/30)^2) = 0.5

So, v = 14.14 m/s

So, ax = 5/30 = 0.167 m/s2

3)

ay = v/t

= 14.14/30

= 0.471 m/s2 <-----answer

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