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CBSE Class 11-science Answered

A car can travel at a max. speed of 180km/hr and can have max. acceleration to be 5m/sec.square and max. retardation can be 3m/sec.square .How fast can it start from rest and come to rest in travelling 300m ?
Asked by sunakshi | 12 May, 2013, 07:20: PM
answered-by-expert Expert Answer
Let us suppose that the car reaches a maximum speed x m/s when it started from rest and then decelerated and moved to rest at final point.
 
Let us suppose the distance travelled when it moved from u=0 to v=x is A
and the distance travelled when it moved from u=x to v=0 is B.
 
So, A+B = 300 .........................1
 
For the first part, A, lets suppose it took time m and for second part B it took time n
Therefore Total time taken = m + n
 
Now for the first part A,
u=0, v=x, a=+5
So, using equation of motion we can write
x2 - 0 = 2*5*A ..............................2
Using another equation of motion we can also write
A= 0m + 5m2/2 ...............................3
 
For path B, we can write similar equation of motion with u=x and v=0 and a=-3
0 - x2 = - 2*3*B ..............................4
Using another equation of motion we can also write
B= x*n - 3n2/2 ...............................5
So from adding equation 2 and 4 we get
5A=3B ...............................6
putting this in equation 1 we get
A= 225/2 and B=375/2.
 
Now putting the value of A in equation 2 we get
x = 15?5
 
Putting the value of B and x in equation 5 we get
n2 - 10?5n + 125 = 0
Solving this quadratic equation would give equal roots at
n=5?5
 
Now putting the value of A in eqn 3 we get
m = 15/?5
 
So total time taken equals to m+n = 15/?5 + 5?5 = 8?5.
 
This is the answer
Answered by | 12 May, 2013, 09:37: PM

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