Question

. Observe the figure and answer the
questions.
С
4 km
5 km
НЕ
A
3 km
B 3 km
D 3 km
on
a
Sachin and Sameer started
motorbike from place A, took the turn
at B, did a task at C, travelled by the
route CD to D and then went on to E.
Altogether, they took one hour for this
journey. Find out the actual distance
traversed by them and the displacement
from A to E. From this, deduce their
speed. What was their velocity from A
to E in the direction AE? Can this
velocity be called average velocity?​

Answer 1

$\huge\sf\red{→solution}$

Given acceleration a=5m/s^2and maximum retardation =10 m/s ^2 and distance between A and B= 1.5 km= 1500 m

considering u= 0 m/s as initially the motorcycle is at rest.

So v=5t., where t is the time taken to accelerate.------(A)

As T is the total time than time taken for retardation by an amount of 10 m/s ^2 be T-t.

From equation of motion we have,

$retardation\ = \frac{final \: veocity - initial}{time}$

In case of retardation or deceleration final velocity will be 0 as the motorcycle will come to rest.

So −v=−10(T−t)-------(B)

Putting the value of A in B we get

−5t=−10T+10t

thus 1/2vt=1500

by=3000

$5t \frac{3t}{2} = 3000$

t^2=400

t=20s

we know, t=3t/2

t=30seconds