car acceleration from rest at constant acceleration 6 m/s2 for some time after which it decelerates atconstant rate 3 m/s2 to come to rest. If the total time is 15 sec. Find maximum velocity reached by car? Pls answer asap
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Answered by
1
Answer:
20 velocity reached by car
Answered by
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=⊥0t →(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=
2
1
×50×15=375 m
∴ average velocity =
t
d
=
15
375
=25 m/s
during deceleration distance travelled, d
1
=
2
1
×(15−5)×50=250m
∴ avg velocity during deceleration =
Δt
d
2
=
(15−5)
250
=25 m/s
solution
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