Question

if velocity of a particle is given by v=10+2t^2 m+s. the average acceleration between 2 and 5s is

Answer 1

check the pic
hint - avg accn = V2-V1/T2-T1
therefore accn=60-18/5-2 =14

Answer 2

Answer:

The average acceleration between 2 and 5s is 15.33m/s².

Explanation:

The average acceleration is given as,

$a=\frac{v_2-v_1}{t_2-t_1}$     (1)

Where,

a=average acceleration

v₁=initial velocity

v₂=final velocity

t₁=initial time

t₂=final time

From the question we have,

$v=10+2t^2\ m/s$     (2)

Initial time(t₁)=2 second

Final time(t₂)=5 second

For t₁=2 second

By placing t₁=2 second in equation (2) we get;

$v_1=10+2(2)^2$

$v_1=10+4$

$v_1=14m/s$    (3)

For t₂=5 second

By placing t₂=5 second in equation (2) we get;

$v_2=10+2(5)^2$

$v_2=10+2\times 25$

$v_2=10+50$

$v_2=60\ m/s$   (4)

By placing all the required values in equation (1) we get;

$a=\frac{60-14}{5-2}$

$a=\frac{46}{3}$

$a=15.33\ m/s^2$

Hence, the average acceleration between 2 and 5s is 15.33m/s².

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