Question

A particle having initial velocity 'u' moves with a constant acceleration 'a' for time 't.'
(a) Find the displacement of the particle in the last 1 second.
(b) Evaluate it for u = 5m/s, a = 2m/s2 and t = 10s

Answer 1

(a). s=ut+0.5×(a×t^2). s=u+a. (b)s=5×10+0.5×2×10^2=50+100=150

Answer 2

Answer:

Step-by-step explanation:

Step 1:

Given Data:

u = 5m/s

$\mathrm{a}=2 \mathrm{m} / \mathrm{s}^{2}$

t = 10 s

Step 2:

Let distance traveled by the particle in last one second be S10

Distance traveled by the particle in nth second of its motion is given by

Sn = u + a/2(2n-1)

Step 3:

Here as the distance traveled in 10th second therefore n = 10  

S10 = 5 + 2/2((2x10)-1)

S10 = 5 + 1(19)

S10 = 24m

Step 4:

The distance traveled by the particle in 10 seconds be S.

We know that for displacement of a particle for a given interval of time  

$\begin{array}{l}{S=u t+1 / 2\left(a t^{2}\right)} \\ {S=(5 \times 10)+1 / 2\left(2 \times(10)^{2}\right)}\end{array}$

S = 50 + 1/2(200)

S = 50 + 100

S = 150 m