Q. 154.4( 23 Votes )

# A car starts from rest and moves along the x-axis with constant acceleration 5 m s–2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

Answer :

As the car moves from rest, where u = 0,

Acceleration, a = 5ms^{-2}

time, t = 8sec

distance(s) =?

From the equation of motion

S_{1} = ut + 1/2 at^{2}

S_{1} = (0x8) + 1/2 x 5 x (8)^{2}

S_{1} = 0 +1/2x5x64

S_{1} = 5x32

S_{1} =160 m

The velocity after 8 sec,

From equation of motion

v = u + at

v = 0 + 5 x 8

v = 40m/s

So, the distance travelled in 4 sec (12 s – 8 s = 4s)

S_{2} = 40 x 4

S_{2} = 160 m

Therefore, total distance travelled by the car = S_{1} + S_{2}

= 160 + 160

= 320 m

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