# 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 | 12th May, 2013, 07:20: PM

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
x^{2} - 0 = 2*5*A ..............................2
Using another equation of motion we can also write
A= 0m + 5m^{2}/2 ...............................3
For path B, we can write similar equation of motion with u=x and v=0 and a=-3
0 - x^{2} = - 2*3*B ..............................4
Using another equation of motion we can also write
B= x*n - 3n^{2}/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
n^{2} - 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

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

x

^{2}- 0 = 2*5*A ..............................2Using another equation of motion we can also write

A= 0m + 5m

^{2}/2 ...............................3For path B, we can write similar equation of motion with u=x and v=0 and a=-3

0 - x

^{2}= - 2*3*B ..............................4Using another equation of motion we can also write

B= x*n - 3n

^{2}/2 ...............................5So 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

n

^{2}- 10?5n + 125 = 0Solving 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 | 12th May, 2013, 09:37: PM

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