Question

A particle accelerates uniformly from 7 ms-1 to 21 ms-1 in 8 s.

How far does it travel in this time?


(b) Lewis is travelling in a car along a straight road. He wonders whether the car is accelerating uniformly.


Lewis estimates that the car takes

5 s to travel a distance of 75 m from A to B, 15 s to travel a distance of 315 m from A to C.

Lewis models the acceleration as a constant a ms-2. He also takes the

speed of the car at A to be u ms-1, as shown in the diagram above.


(i) By considering the motion from A to B, show that

75 = 5u + 12.5a


(ii) Find a second equation involving u and a.


(iii) Hence find the value of u and show that a = 1.2

Thank you very much in advance if you take the time to help me out!
​

Answer 1

9514 1404 393

Answer:

A) 112 m

Bi) 75 = 5u +12.5a

Bii) 21 = u +12.5a

Biii) a = 0.6; u = 13.5

Explanation:

A) The average speed is ...

(7 m/s +21 m/s)/2 = 14 m/s

At that speed, in 8 seconds the car travels ...

(14 m/s)(8 s) = 112 m

__

B) Using similar reasoning, we note Lewis's average speed from A to B is ...

(75 m)/(5 s) = 15 m/s

(i) If acceleration is uniform, this is the speed at the middle of the interval. It will be the sum of the initial speed (u) and the increase due to acceleration over half the interval (2.5a). So, the relation we've described is ...

15 = u +2.5a

Multiplying by 5 gives the relation we're to show:

75 = 5u +12.5a

__

(ii) The average speed in the second interval is ...

(315 m)/(15 s) = 21 m/s

The midpoint of that interval is 5+(15/2) = 12.5 s after the start of timing. Then the second equation can be ...

21 = u + 12.5a

__

(iii) The solution to these two equations is ...

(21) -(15) = (u +12.5a) -(u +2.5a) . . . . . subtract the 1st equation from the 2nd

6 = 10a

a = 0.6 . . . . not 1.2

Then the value of u is ...

u = 15 -2.5(0.6)

u = 13.5

_____

The attached graph shows the velocity vs time curve and a table of velocity and distance values. The table values match those in the problem statement, even if the equation value for acceleration does not match the problem statement.

f(t) is the velocity as a function of time

d(x) is the distance as a function of time

You will note that d(5) = 75, and d(5+15) = d(20) = 75 +315 = 390.