# initial acceleration .Watch

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#1
Hey guys .
im struggling sorting out the below question especially on part a,c,d . i've managed to do only part b .
any help would be much appreciated as i got stuck so bad on this one 0
4 years ago
#2
a) Initial acceleration is equal to the gradient of the graph (due to acceleration being change in velocity over time).
c) You would use the same method as b, but just do it for the next 2 seconds as opposed to the first two.
0
4 years ago
#3
(Original post by Alen.m)
Hey guys .
im struggling sorting out the below question especially on part a,c,d . i've managed to do only part b .
any help would be much appreciated as i got stuck so bad on this one Part a: Acceleration = rate of change of velocity
How could you get a 'rate of change' of something from a graph?

Part c: Strange, you managed to do part b but didn't know how to deal with part c. They are the same in principle Part d: Average speed = Change in distance/Time taken
Can you see how to go from here?

Hope this helps!
0
#4
i have calculated the gradient of the graph for part a to get initial acceleration but i keep getting wrong answer compare to the answer the book said which is 5ms*-2 .
for part c i've calculated the area under the line till 4s but my answer is again different with the book .
should i take away the distance i've measured for part b and c which d1 and d2 and divide by the time of distances which t1 and t2 in order to get the average speed?
thanks again .
0
4 years ago
#5
(Original post by Alen.m)
i have calculated the gradient of the graph for part a to get initial acceleration but i keep getting wrong answer compare to the answer the book said which is 5ms*-2 .
for part c i've calculated the area under the line till 4s but my answer is again different with the book .
should i take away the distance i've measured for part b and c which d1 and d2 and divide by the time of distances which t1 and t2 in order to get the average speed?
thanks again .
How did you calculate the gradient? Because if you drew the tangent of the curve at (0,0) then you would see that the gradient is 5ms^-2, which is the answer.

For part c, the question asked 'in the next 2s' so the area wanted is from 2 to 4 under the curve.

The last part will be the sum of two distances divided by total time taken (4 secs).
0
#6
thanks so much i got it now for part A as the tangent i drawn was absolutely wrong . just to remind me the area for part c would be an estimate or does it have to be accurate?can't get to the answer 18 0
4 years ago
#7
(Original post by Alen.m)
thanks so much i got it now for part A as the tangent i drawn was absolutely wrong . just to remind me the area for part c would be an estimate or does it have to be accurate?can't get to the answer 18 You're welcome The area would be an estimate only (as you don't have the exact equation of the curve to determine its area), but you can see it consists of just more than 7 'rectangular blocks', each has an area of 2.5, so the total area would be approximately 18. Hope this helps!
0
#8
yeah it helped a lot . thanks in advance
0
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