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 rest?

320 m

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160 m

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480 m

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720 m

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Solution

The correct option is A 320 m
$s_{1} = u t + (1 / 2) a t^{2}$
$= 0 + \frac{1}{2} \times 5 \times 8^{2}$
$= 160 m$ .
The velocity attained in 8 seconds is given by
$v = u + a t$
$= 0 + 5 \times 8 = 40 m / s$ .
The distance traveled in the next 4 seconds is
$s_{2} = 40 \times 4 = 160 m$ .
So, the total distance traveled by the car is $s_{1} + s_{2} = 160 + 160 = 320 m$

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

The correct option is A 320 ms1=ut+(1/2)at2 =0+12×5×82 =160 m.The velocity attained in 8 seconds is given byv=u+at =0+5×8=40 m/s.The distance traveled in the next 4 seconds is s2=40×4=160 m.So, the total distance traveled by the car is s1+s2=160+160=320 m

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 rest?