Question

The acceleration of a cart started at t = 0, varies with time as shown in figure. The car starts from rest. Then the distance travelled by cart in 30 seconds is:​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

In 1

st

10 sec S

1

=ut+1/2 at

2

⇒0+(1/2×5×10

2

)=250 ft

At 10 sec v=u+at=0+5×10=50 ft/sec

∴ From 10 to 20 sec (Δt=20−10=10 sec) it moves with uniform velocity 50 ft/sec,

Distance S

2

=50×10=500 ft

Between 20 sec to 30 sec acceleration is constant i.e. −5 ft/s

2

. At 20 sec velocity is 50 ft/sec.

t=30−20=10 s

S

3

=ut+1/2 at

2

=50×10+(1/2)(−5)(10)

2

=250 m

Total distance travelled is 30 sec=S

1

+S

2

+S

3

=250+500+250=1000 ft.