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Question

A particle starts from rest, accelerates at $$2\ m/s^{2}$$ for $$10s$$ and then goes for constant speed for $$30s$$ and then decelerates at $$4\ m/s^{2}$$ till it stops. What is the distance travelled by it?


A
750 m
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B
800 m
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C
700 m
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D
850 m
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Solution

The correct option is A $$750\ m$$
A)
Initially particle starts from rest. So$$ u=0 \ and \ a=2m/s^2 and t=10 sec$$ and formula for this case is: 
$$s=ut+(1/2)a(t^2)$$ and using this 
$$s = 0*10 + (1/2)*2*(10^2) = 100m$$  
and using the given information and this formula 
v = u + at 
we get the value of final velocity and on substituting all values we get:
0+2*10 = 20m/sec 
Given that we have constant speed implies initial velocity is the same as the final velocity. 

v=u and using this information and the formula $$v=u+at  $$ 
we get a=0 as t can’ t be zero. and $$t=30 sec$$ (given). And from part $$u=20m/sec$$

Now to get the value of distance in this part we use this formula :
$$ s=ut+(1/2)a(t^2)$$ and using this we get:
$$ 20*30+(1/2)*0*(30^2) = 600m .$$

Now $$ a=(-)4m/sec^2$$
as particle is decelerating and given statement is it decelerates till is stops  implies that final velocity is zero. And from above part $$u=20m/sec$$ and to get the value of distance we use this formula:
$$ v^2-u^2 = 2as$$ and using this we get: 
$$0-20^2 = 2*(-4)*s =>  s =400/8 = 50m.$$
 So using all the part info we get total distance travelled: 100+600+50 = 750

Physics

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