## Q6 A motorcyclist heading east through a small Iowa city accelerates after he passes the signpost marking the city limi…

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Q6 A motorcyclist heading east through a small Iowa city accelerates after he passes the signpost marking the city limits. His acceleration is a constant 4.0 m/S2. At time t = 0 he is 5.0 m east of the signpost, moving east at 15 m/s. (a) find his position and velocity at time t = 2.0 s. (b) Where is the motorcyclist when his velocity is 25 m/s?

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6 months 2023-04-07T05:49:11+00:00 1 Answer 0 views Teacher 0

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Given :

constant acceleration of motorcyclist, a = 4 m/s�
position of motorcyclist at time t = 0 , x? = 5 m east of signpost
initial velocity of motorcyclist, v? = 15 m/s

To find :

position of motorcyclist at t = 2 s , x = ?
velocity of motorcyclist a t t = 2 s , v? = ?
position of motorcyclist when its velocity will be 25 m/s , x’ = ?

Knowledge required :

First equation of kinematics

v? = v? + a t

Second equation of kinematics

x = x? + v? t + 1/2 a t�

Third equation of kinematics

v?� = v?� + 2 a ( x – x? )

[ where v? is final velocity (velocity acquired after a time t ), v? is initial velocity , t is time taken , a is constant acceleration , x is distance covered (after a time t) , and x? is initial position of any body ]

Solution :

Calculating velocity motorcyclist at a time t = 2 sec ; v? = ?

Using first equation of kinematics

? v? = v? + a t

? v? = ( 15 ) + ( 4 ) ( 2 )

? v? = 15 + 8

? v? = 23 m/s

therefore,

velocity of motorcyclist after 2 seconds will be 23 m/s .

Calculating position of motorcyclist at time t = 2 sec ; x = ?

Using second equation of kinematics

? x = x? + v? t + 1/2 a t�

? x = ( 5 ) + ( 15 ) ( 2 ) + 1/2 ( 4 ) ( 2 )�

? x = 5 + 30 + 8

? x = 43 m

therefore,

position of motorcyclist after 2 seconds will be 43 m east from the signpost.

Calculating position of motorcyclist when its velocity will be v = 25 m/s ; x’ = ?

Using third equation of kinematics

? v� = v?� + 2 a ( x’ – x? )

? ( 25 )� = ( 15 )� + 2 ( 4 ) ( x’ – 5 )

? 625 – 225 = 8 ( x’ – 5 )

? 400 / 8 = x’ – 5

? x’ = 50 + 5

? x’ = 55 m

Therefore,

position of motorcyclist when its velocity becomes 25 m/s will be 55 m east from the signpost.