A dog initially running at a speed of 2 ms^-1, accelerates at a constant rate of 1.5 ms^-2 for 5 seconds. Calculate
(i) the distance covered by the dog
from its initial position
(ii) the finial velocity of dog
[ Ans 28.75 m, 9.75 ms^-1]
Answers
Answer:
It is given that,
u = 2m/s
a = 1.5m/s^2
t = 5 sec
(a) Using 2nd equation of motion, i.e., s = ut + 1/2 x a x t
s = 2 x 5 + 1/2 x 1.5 x 5
= 10 + 0.5 x 1.5 x 5
= 10 + 3.75
= 13.75 m
(b) Using 1st equation of motion,
i.e., v = u + at
v = 2 + 1.5 x 5
= 2 + 7.5
= 9.5 m/s
Ans (a) The distance covered by the dog is 13.75 metre.
Ans (b) The final velocity of the dog is 9.5 m/s.
Explanation:
As per Newton's equations of motion,
S=u×t+1/2×a×t²
where S is the distance from the initial position ,
u is the initial velocity and 'v' is the final velocity and
'a' is the acceleration and 't' is the time taken.
if we substitute the values for this equation, we get the ans to the 1st question,
S=2×5+1/2×1.5×5²
=10+1.5×12.5
=10+18.75
=28.75 m
and if we want the final velocity, we know that acceleration is the difference between final and initial velocities divided by time for the change in velocity,
that is a=v-u÷t
1.5=v-2÷5
if we bring the 5 to the other side,
1.5×5=v-2
7.5=v-2
∴v=7.5+2
=9.5