A car accelerates form rest at a constant rate fot t seconds and covers distance x . The distance covered by it in the next t seconds is

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Solution

Dear Student,

Initial speed of the particle, u = 0

For the first t seconds, using the equation of motion
$S = u t + \frac{1}{2} a t^{2} \Rightarrow x = 0 + \frac{1}{2} a t^{2} \Rightarrow a = \frac{2 x}{t^{2}}$

Distance covered in another t seconds,
$x' = (u t + \frac{1}{2} a {(2 t)}^{2}) - (u t + \frac{1}{2} a {(t)}^{2}) \Rightarrow x' = (\frac{1}{2} a 4 t^{2}) - (\frac{1}{2} a t^{2}) = \frac{3 a t^{2}}{2} \Rightarrow x' = \frac{2 x \times 3 t^{2}}{2 \times t^{2}} = 3 x$

Regards,

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A car accelerates form rest at a constant rate fot t seconds and covers distance x . The distance covered by it in the next t seconds is

Dear Student, Initial speed of the particle, u = 0 For the first t seconds, using the equation of motion S=ut+12at2⇒x=0+12at2⇒a=2xt2 Distance covered in another t seconds, x'=ut+12a2t2-ut+12at2⇒x'=12a4t2-12at2=3at22⇒x'=2x×3t22×t2=3x Regards,