Question

A car starts from rest, attains a velocity of 36 km h'
with an acceleration of 0.2 m stravels 9 km with
this uniform velocity and then comes to halt with
a uniform deceleration of 0.1 ms. The total time
of travel of the car is
(a) 1050 s (b) 1000 s (c) 950 s (d) 900 s​

Answer 1

Explanation:

Given

v=36kmh

−1

=

3600

36000

=10ms

−1

s=9km=9000m

This motion is in three phases.

[Case: 1]

with acceleration, a=0.2ms

−2

and intial velocity , u=0

Thus using equation v=u+at

10=0+0.2×t

So, t=50 s

[Case: 2]

uniform motion with velocity ,v=10

and distance, s=9000

Thus time , t=

v

s

=

10

9000

=900 s

[Case: 3]

With acceleration, a=−0.1ms

−2

u=10ms

−1

and v=0

Thus using equation v=u+at

0=10−0.1×t

So, t=

0.1

10

=100 s

Therefore

Total time = 50s+900s+100s=1050s

Answer 2

Explanation:

total time will be 1000s

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