If a car rest accelerated uniformly to a speed of 144 -km-hour- in 20 second it covers a distance-1440 m400 m2880 m25 m

Question

Toppr Admin · Accepted Answer

-                            -u-0-v-144km-hour-144-xD7-518m-sec-40m-secv-u-at-x21D2-a-v-x2212-ut-40-x2212-020-2m-sec2-x2234-s-ut-12at2-12-xD7-2-xD7-20-2-400m-

If a car at rest accelerated uniformly to a speed of $144 k m / h o u r$ in $20 s e c o n d$ it covers a distance:
$1440 m$
$400 m$
$2880 m$
$25 m$

$u = 0, v = 144 k m / h o u r = 144 \times \frac{5}{18} m / s e c = 40 m / s e c$
$v = u + a t$
$\Rightarrow a = \frac{v - u}{t} = \frac{40 - 0}{20} = 2 m / {s e c}^{2}$
$∴ s = u t + \frac{1}{2} a t^{2}$
$= \frac{1}{2} \times 2 \times {(20)}^{2} = 400 m$

If a car at rest accelerated uniformly to a speed of 144km/hour in 20second it covers a distance:1440m400m2880m25m

u=0,v=144km/hour=144×518m/sec=40m/secv=u+at⇒a=v−ut=40−020=2m/sec2∴s=ut+12at2=12×2×(20)2=400m

If a car at rest accelerated uniformly to a speed of $144 k m / h o u r$ in $20 s e c o n d$ it covers a distance:
$1440 m$
$400 m$
$2880 m$
$25 m$

$u = 0, v = 144 k m / h o u r = 144 \times \frac{5}{18} m / s e c = 40 m / s e c$
$v = u + a t$
$\Rightarrow a = \frac{v - u}{t} = \frac{40 - 0}{20} = 2 m / {s e c}^{2}$
$∴ s = u t + \frac{1}{2} a t^{2}$
$= \frac{1}{2} \times 2 \times {(20)}^{2} = 400 m$