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Grade 10Mechanics

The acceleration of a cart started at t = 0, varies with time as shown in figure (3-E2). Find the distance travelled in 30 seconds and draw the position-time graph.

Hrishant Goswami

12 Years agoGrade 10

1 Answer

Jitender Pal

12 Years ago

Sol. In 1st 10 sec S1 = ut + 1/2 at2 ⇒ 0 + (1/2 x 5 x 102 ) = 250 ft. At 10 sec v = u + at = 0 + 5 × 10 = 50 ft/sec. ∴ From 10 to 20 sec (∆t = 20 – 10 = 10 sec) it moves with uniform velocity 50 ft/sec, Distance S2 = 50 × 10 = 500 ft Between 20 sec to 30 sec acceleration is constant i.e. –5 ft/s2. At 20 sec velocity is 50 ft/sec. t = 30 – 20 = 10 s S3 = ut + 1/2 at2 = 50 x 10 + (1/2) (-5) (10)2 = 250 m Total distance travelled is 30 sec = S1 + S2 + S3 = 250 + 500 + 250 = 1000 ft.

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