Question

A car accelerates from rest at a constant rate 10 m per second square for some time after which it decelerates at constant rate 5 m per second square to come to rest. If total time elapsed is 15s.The time (in s) during which the particle is in the condition of deceleration

Answer 1

Answer:

Explanation:

Suppose car acceleration for time t. So it decelerate by time (15−t)s.

Initially at rest, u=0 and come to velocity v in time t! use from Kinematics  

for constant accim, (a=10m/m^2)

⇒ v=⊥Ot →(1)

⊥ & (2) gives,

⊥Ot=5(15−t) ⇒ t=5 s.

For constant deceleration (a=−5m/s)

⇒ o=v−5(15−t)

⇒ v=5(15−t) →(2)

so, v=⊥ot=50 m/s

construct a v−t graph from obtained results.

From graph

(i) max velocity =50 m/s

(ii) net distance travelled =area ot vt graph

d=1/2*50*15=375 m

Therefore average velocity = d/t=375/15=25 m/s

During deceleration the distance travelled. d1= 1/2(15-5)*50

Therefore average velocity during deceleration=d2/Δt=250/(15-5)

=25m/s

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