# Mechanics help?

May 14, 2018

i)Firstly you should sketch velocity time graph for this question with considering maximum velocity $v$ and mentioning time on x-axis

As acceleration=gradient in velocity time graph
Or you can say
Acceleration=rise/run

Thus you can find velocity easily that will be $3 = \frac{v}{10}$

So $v = 20$m/s

And as they have asked you to find distance and in velocity time graph $\mathrm{di} s \tan c e = a r e a u n \mathrm{de} r t h e g r a p h$

Find area under the graph look at you rough sketch it will be the trapezim area

$\frac{1}{2} \cdot$(length of parallel side)$\cdot h$
height is the velocity you found above as $30$m/s

$\frac{1}{2} \cdot \left(40 + 30\right) \cdot 30$

So you distance is $1050 m$

ii)
(ii) Let the car comes to rest after seconds at B. (Assume)
And assume A is point (0,0) on graph

Total distance travelled by car=area of trapezium
$1500 = \frac{1}{2} \cdot \left(30\right) \left(30 + t\right)$
$100 = 30 + t$
$t = 70$
The motor cycle starts 20s later
Thus,
total time taken by car from AtoB=$70 - 20$
That is $50$seconds

Total distance travelled by motorcycle=area of trapezim

That is $1500 = \frac{1}{2} \cdot V \cdot \left(30 + 50\right)$

$1500 = 40 V$
$v = 37.5$m/s

maximum speed reached by motorcycle is, V 37.5 ms

motorcycle travels at constant speed for 30 seconds, therefore total time taken for acceleration and deceleration=20 seconds.

Let time taken for acceleration is t seconds, time taken for deceleration is (20-t)seconds given that., acceleration=3*deceleration

$\left(\frac{37.5}{t}\right) = 3 \cdot \left(\frac{37.5}{20 - t}\right)$

$20 - t = 3 t$

$4 t = 20$

$t = 5$s

Acceleration
a=$\frac{37.5}{5}$

$a = 7.5$m/s^2

For iii) displacement time graph would be like this I am writing coordinates you can make it

$\left(0 , 0\right)$
$\left(10 , 150\right)$
$\left(40 , 1050\right)$
$\left(70 , 1500\right)$