Question

along
car
5
8.
r starts from
n rest and most ale
the x-axis with
constant acceleration
5 m/s² for 8 seconds. If it then contities
continues with constant velocity what
distance will the car cover in seconds
since it started from the rest​

Answer 1

v=u+at

v=0+5m/s^2*8

v=40m/s

constant velocity=40m/s

distance covered in the first 8 sec = S=1/2a*t^2=160m.

distance covered in the last 4 sec=const. speed*time=40m/s*4=160m

total distance covered in 12 sec=160+160=320m ans.

ans=320m

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________________________________________________

lets divide the qstn into 2 parts..

(1) the distance it travelled in the frst 8 sec

(2) the distance it travld inte next 4 secs..that is in total 12 sec.

(1) given,

 u=0

a=5m/s^2

t=8s

using the eqn s=ut+1/2 at2,

s=0+(1/2)*5*(8^2)

s=160m..................this s d distnce travelled in 8 sec.

now while considering the 2nd part the final velocity in the frst becomes the initial in the sec.

find the final velocity. v=u+at..(not sure abt dis eqn....sry.)

u get v=40m/s.

in the secnd prt it is said that the body moves with uniform velocity. ie a=0

using the eqn s=ut+1/2at2,

u get s=(40*4)+0

s=160m.

thrfore the total distance travelled equals 160+160

=320m.

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