Question

The acceleration time graph (at) of a particle
moving with initial
speed of 45 m/s
below.
the velocity of a body at t = 20s is
om
is shown
a cms 2)
- 2​

Answer 1

Explanation:

Given : V i = 2m/s at t=0 s

Let V f be the velocity of the particle at t=4 s

Area under the a-t graph gives the change in velocity of the particle during that period.

∴ ΔV = V f − V i

= 1 ×4×4 = 8

2

OR V f −2=8

⟹V f = 10m/s

Answer 2

Answer:

Acceleration  be 225cms⁻².

Explanation:

v= final velocity=0ms⁻¹ at t=20s

u=initial velocity=45ms⁻¹ at t= 0s

Acceleration=a

• V=u+at
• 0= 45+a*20

        $v=u+at\\ 0=45+a*20\\-45=20a\\\\a=\frac{-45}{20} \\\\ = 2.25ms^{-2} \\\\= 225cms^{-2}$