Question

An object accelerates from rest to a velocity 27.5
ms^-1 in 10s. Find the distance covered by the object
during the next 10s
(1)412.5m(correct answer )
(2) 137.5m
(3) 550m
(4) 275m
need explanation friends....​

Answer 1

Explanation:

u=0

v=27.5m/s

a= unknown

t=10secs .

using first equation of motion

v=u+at

27.5=0+10(a)

a=2.75m/s^2

we have got a now . distance for next 10sec :

a=2.75m/s^2

u=27.5m/s [v becomes u for considering next ten second]

t=10s

s=unknown

using

s=ut+at^2/2

s= 275+ (2.75)(100)/2

s= 275+ 275/2

s=275+137.5

s= 412.5 m

so distance travelled in next ten seconds is 412.5m

hope this helps.

all the best for your future

endeavours

jai hind

Answer 2

Explanation:

click the attachment mate