A car starts from rest and moves along the x-axis with constant acceleration `5 ms^(-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 ?
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Given, the car starts from rest so its initial velocity u = 0 Acceleration, (a)= `5ms^(-2)` and time (t) = 8s From first equation of motion, v = u + at On putting a = `5ms^(-2)` and t = 8 s in above equation, we gel `v = 0 + 5 xx 8 = 40ms^(-1)` So, final velocity v is `40ms^(-1)` . Again, from second equation of motion. s = ut `+1/2at^(2)` On putting t = 8s and a = `5ms^(-2)` in above equation we get `s= 0 xx 8 +1/2xx5xx(8)^(2) = 1/2 xx 5 xx 64 = 5 xx 32 = 160 m` So, the distance covered in 8 s is 160m. Given, total time t = 12s. After 8s, the car continues with constant velocity i.e., the car will move with a velocity of `40ms^(-1)` So, ramaining time t' = `12s - 8s = 4s` The distance covered in the last 4s(s) = Velocity `xx` Time [`because` Distance = Velocity `xx` Time] = `40xx4` = 160m [We have used the direct formual because after 8s, car is moving with constant velocity i.e., zero acceleration] `:.` Total distance travelled in 12 s from the start D = `s + s = 160 + 160 = 320m`
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