If a car at rest accelerates uniformly to a speed of 144 km /hour in 20 second it covers a distance of

Question

If a car at rest accelerates uniformly to a speed of 144 km /hour in 20 second it covers a distance of (A) 400 m (B) 1440 m (C) 2880m (D) 25m

Tardigrade Admin · Accepted Answer

Initial velocity u=0, Final velocity =144 km / h =40 m / s and time =20 sec. Using v=u+a t ⇒ a=v / t=2 m / s 2 Again, s=u t+(1/2) a t2 =(1/2) × 2 ×(20)2=400 m.

Q. If a car at rest accelerates uniformly to a speed of $144 km$ /hour in $20$ second it covers a distance of

400 m

1440 m

2880m

25m

Initial velocity $u = 0$ ,
Final velocity $= 144 km / h = 40 m / s$
and time $= 20 sec$ .
Using $v = u + a t$
$\Rightarrow a = v / t = 2 m / s^{2}$
Again, $s = u t + \frac{1}{2} a t^{2}$
$= \frac{1}{2} \times 2 \times (20)^{2} = 400 m$ .

Q. If a car at rest accelerates uniformly to a speed of 144km /hour in 20 second it covers a distance of

400 m

1440 m

2880m

25m

Initial velocity u=0, Final velocity =144km/h=40m/s and time =20sec. Using v=u+at ⇒a=v/t=2m/s2 Again, s=ut+21​at2 =21​×2×(20)2=400m.

Q. If a car at rest accelerates uniformly to a speed of $144 km$ /hour in $20$ second it covers a distance of

Initial velocity $u = 0$ ,
Final velocity $= 144 km / h = 40 m / s$
and time $= 20 sec$ .
Using $v = u + a t$
$\Rightarrow a = v / t = 2 m / s^{2}$
Again, $s = u t + \frac{1}{2} a t^{2}$
$= \frac{1}{2} \times 2 \times (20)^{2} = 400 m$ .