Question

2.
Observe the figure and answer the
questions.
С
4 km
5 km
A 3 km
on
B 3 km D 3 km
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?
From​

Answer 1

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,

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

t^2=400

t=20s

we know, t=3t/2

t=30seconds

Explanation:

the answer is 30 second.....