Question

An automobile starting from rest speeds up to 1.2m/s^2, maintaining the speed for sometime and finally come to rest with a deceleration of 1.5m/s^2. If the total distance traveled is 300m, find the total time traveled.

Answer 1

Answer:

If t

1

be the time for which the car accelerates at the rate f from rest, the distance traveled in time t

1

is

s=s

1

=0(t

1

)+

2

1

ft

1

2

=

2

1

ft

1

2

....(1) (using formula

s=ut+

2

1

at

2

)

and the velocity at time t

1

is v

1

=ft

1

So, the distance traveled in time t will be s

2

=v

1

t=ft

1

t ...(2)

If t

2

be the time for which the car decelerates at the rate f/2 to come rest, the distance traveled in time t

2

is given by,

0

2

−v

1

2

=2(−f/2)s

3

(using formula v

2

−u

2

=2as)

or s

3

=(ft

1

)

2

/f=ft

1

2

...(3)

using (1), s

3

=2s

1

=2s

Given, s

1

+s

2

+s

3

=15s

or s+ft

1

t+2s=15s

or ftt

1

=12s

or 12s=ftt

1

..(4)

(4)/(1)⇒

s

12s

=

(1/2)ft

1

2

ftt

1

or t

1

=t/6

From (1), s=(1/2)f(t/6)

2

=ft

2

/72

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