Question #0f8cd

2 Answers
Narad T.
Sep 6, 2017

Tje answer to part (c) is 40.5kmh−1

Explanation:

u=54kmh−1=54⋅10003600=15ms−1

v=36kmh−1=36⋅10003600=10ms1

The acceleration is

a=v−ut=10−152=−52=−2.5ms−2

The distance travelled is

s=v2−u22a=100−225−2.5⋅2=−125−5=25m

The speed of the car when t=1.5s

v1=u+at=15−2.5⋅1.5=15−3.75=11.25ms−1

=11.25⋅36001000=40.5kmh−1

Mr. Raj
Sep 6, 2017

v=11.25.ms−1

Explanation:

From part a, you know what the acceleration is −2.5 ms−2
For part c, first list out the values you do know:

v0=54kmh−1=15ms−1
a=−2.5 ms−2
t=1.5.s
v=?(the v given in the question is after 2 secs, so it can't be used)

(we do know the distance the car travels in 2 seconds, but this is not useful in this case because we are looking at 1.5 seconds)

Using the values we do know, we can determine the equation that we should be using. Choose out of the equations of motion.
We can see that the only one with v0,a,t and v is the first one:

v=v0+at

Insert the values we know, and solve for v:

v=(15)+(−2.5)(1.5)

⇒v=11.25.ms−1=40.5.kmh−1

Just as a side note: When doing calculations like these, make sure that you use constant units, try to avoid changing between unit midway through doing questions. I like to work with the SI units because most questions would want it in that.

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