Question

Please solve the above question...

Answer 1

i) from t=0 to t=4, body is accelerating

from t=4 to t=6, body is decelerating

ii) from 0 to 4 seconds

$a = \frac{change \: in \: velocity}{time \: taken}$

$a = \frac{2}{4} = \frac{1}{2} m {s}^{ - 2}$

using third equation

$v {}^{2} = {u}^{2} + 2as$

$4 = 0 + 2 \times \frac{1}{2} s$

$s = 4 \: m$

from 4 to 6 seconds

acceleration is in opposite direction, so we'll be taking its sign at negative

$a = - \frac{2}{2} = - 1 \: m {s}^{ - 2}$

using third equation of motion

$v {}^{2} = u {}^{2} + 2as$

$0 = 4 + 2( - 1)s$

$s = 2 \: m$

total displacement= 4+2= 6m

iii) yes

iv) distance travelled from 0 to 4 s= 4m

distance travelled from 4 to 6 s= 2m

$ratio = \frac{4}{2} = \frac{2}{1}$

v) acceleration and retardation have been calculated.