Question

v=t+1.find speed at the end of 4sec. If initially body is at rest or origin.
Please reply fast it's urgent. ​

Answer 1

Answer:

u=0,a (uniform acceleration),t=n

v (final velocity after n seconds)

v=0+an⟹v=an⟶(1)

sn (distance travelled in n seconds)=0+21an2

sn−2=21a(n−2)2

Distance travelled in least 2 seconds       =sn−sn−2=21a[n2−(n−2)2]=21a[n2−n2−4+4x]=2an−2a⟶(2)

From equation 1 and 2,

x=2a(n−1)x=u2v(n−1)

Answer 2

Explanation:

u=0,a (uniform acceleration),t=n

v (final velocity after n seconds)

v=0+an⟹v=an⟶(1)

sn (distance travelled in n seconds)=0+21an2

sn−2=21a(n−2)2

Distance travelled in least 2 seconds       =sn−sn−2=21a[n2−(n−2)2]=21a[n2−n2−4+4x]=2an−2a⟶(2)

From equation 1 and 2,

x=2a(n−1)x=u2v(n−1)

IF YOU LIKE MY ANSWER SO PLEASE MARK me AS BRAINLIEST