Question

An object P travels from A to B in a time of 80 s. The diagram shows the graph of v against t, where
vm s−1 is the velocity ofP at time ts after leaving A. The graph consists of straight line segments for
the intervals 0 ≤ t ≤ 10 and 30 ≤ t ≤ 80, and a curved section whose equation is v = −0.01t
2 +0.5t−1
for 10 ≤ t ≤ 30. Find
i) the maximum velocity of P
ii) the distance AB

Answer 1

we need the equation of the straight line segments for
0 ≤ t ≤ 10
and
30 ≤ t ≤ 80

also, is it
10 ≤ t ≤ 30 \( v = −0.01t^2 +0.5t−1 \)

Answer 2

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s08_qp_4.pdf

Answer 3

Question 7

Answer 4

The max velocity we see from the graph occurs in the interval 10 ≤ t ≤ 30
we can use calculus to find the max t and corresponding v
\[ \frac{dv}{dt} = - \frac{t}{50} + \frac{1}{2}= 0 \]

Answer 5

okay i understand this step well but question 7 ii

Answer 6

how am i gonna workout question 7 ii)

Answer 7

the total distance is
\[ \int v \ dt \]
i.e. the area under the curve.
the first region is a triangle with base= 10. Its height (assuming continuity... based on the graph) is v= −0.01t^2 +0.5t−1 evaluated at t=10
similarly for the right most triangle (base 50, height v at t= 30)

the middle region is the integral
\[ \int_{10}^{30} −0.01t^2 +0.5t−1 \ dt \]

Answer 8

thank you :D

Answer 9

yw