Question

A motor bike covers 38.0 m in 12 s while slowing down to a final speed of 4.75 m/s. Find its original speed.
(Assume constant acceleration),​

Answer 1

Answer:

(a) Here,

x

f

−x

i

=

2

1

(v

i

−v

f

)t becomes 40.0m=

2

1

(v

i

+280m/s)(8.50s)

Which yields v

i

=6.61m/s

(b) Similarly,

a=

t

v

f

−v

i

=

8.50s

2.80m/s−6.61m/s

=−0.448m/s

2

.

I HOPE IT IS HELPFUL TO YOU

Answer 2

Given : A motor bike covers 38.0 m in 12 s while slowing down to a final speed of 4.75 m/s.  

(Assume constant acceleration),​

To Find :   original speed.

Solution:

S = 38 m

u = ?

v = 4.75 m/s

t = 12 s  

a =  ?

S = ut + (1/2)at²

=> 38  = 12u  + (1/2)a(12)²

=> 38 = 12u  + 72a

=> 19 = 6u  + 36a

v = u + at

=> 4.75  = u  + 12a

19 = 6u  + 36a

4.75  = u  + 12a  => 14.25  = 3u  + 36a  

=>  4.75 = 3u  

=> u = 4.75/3  m/s

Looks mistake in data  

as u should be more than v as per question  as slowing down mentioned

or direct formula

S = {( v + u)/2 }  * t

=> 38  = ( 4.75 + u ) * 6

=> 6.33 =   4.75 + u

=> u = 1.5833  m/s

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