Question

during launching the space rocket accelerates uniformly from rest to 200 m per second in 5 seconds and then Travels at constant speed for 4 seconds how far does the rocket travels in 9 seconds

Answer 1

v= at
a= 200/5= 40

$s1 = \frac{1}{2} a {t}^{2}$
s1= 500m
s2= 200 × 4= 800m
Total dist= 1300 m

Answer 2

The primary formula for this is V= U+AT for first 5 second where V and U is final and initial velocity respectively. A is the acceleration and T is time. So according to the formula 200=0 + A.5 i.e. A=40.

So without using any formula the rocket travels at (200 × 4) For the last 4 seconds that is 800m. And for the first five seconds it travels (40+80+120+160+200)m that is 600m. So total distance travelled is 600+800 m=1400m