A bus starts moving from rest with constant
acceleration of 1 m/s2. A man standing 48 m
behind the bus starts running towards bus
simultaneously with speed 10 m/s. The man can
catch the bus
(1) Only once at t = 8 s
(2) Only once at t = 10 s
(3) Twice at t = 8 s and t = 12 s
(4) Twice at t, = 6 s and t = 12 s
Answers
Answered by
1
Answer:
Acceleration of bus, a=1ms
−2
Assume man catches the train in timet.
Displacement by man in time t s
1
=vt=10t
Displacement by bus in time t, s
2
=ut+
2
1
at
2
=
2
t
2
Separation between man and bus is 48m
s
1
=s
2
+48
10t=
2
t
2
+48
t
2
−20t+96=0
t=
2
20±
20
2
−4×96
t=8and12
In first attempt he caches the bus
Hence, minimum time he catches the bus is 8sec
Answered by
4
Answer:
let time taken by man = time taken by bus =x
for bus:
u=0
a=1m/s²
t=x
S=ut+1/2at² =o×x +1/2×1×x² =x²/2
for man :
v=10m/s
t=x
S=v×t = 10x
distance travelled by man - distance travelled by bus =48
so, 10x-x²/2 =48
20x-x²/2=48
20x-x²=48×2
=96
20x -x²-96 =0
x²-20x+96=0
x²-12x-8x+96=0
x(x-8)-12(x-8)=0
(x-12)(x-8)=0
x=12 or x=8
minimum time = 8sec
Explanation:
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