Question

) A truck starts from rest and attains a velocity of 40m/s in 5 sec.

calculate acceleration of the truck and distance covered by truck

during this time.​

Answer 1

Answer:

8 m/s²

Step-by-step explanation:

u = 0 m/s

v = 40 m/s

t = 5 sec

From first equation of motion, we get

v = u+at

=> 40 = 0+5a

=> 5a = 40

=> a = 40/5

. ' . a = 8 m/s²

Answer 2

Given:-

initial velocity(u)=0

final velocity(v)=40m/s

time(t)=5 seconds

To find:-

i.acceleration(a)

ii.distance covered(s)

Solution:-

i.a=v-u/t

a=40-0/5

a=40/5

a=8m/s²

ii.v²-u²=2as

(40)²-0=2×8×s

1600=16s

s=1600/16

s=100m

Thus acceleration is 8m/s² and the dustance covered is 100m.