Answer:
Given a-t graph:
(i) Between 0 to 10s, a is constant, i.e., \[5ft/{{s}^{2}}.\]
\[{{t}_{2}}=?\]In the first 10s, distance covered,
\[{{S}_{1}}=ut6+\frac{1}{2}a{{t}^{2}}=0+\frac{1}{2}\times 5\times {{10}^{2}}=250ft\]c
At 10s, velocity acquired,\[\Rightarrow \]
\[=0+5\times 10=50m/s\]
(ii) Between 10s to 20s [i.e.,\[\Delta t=20-10=10s]\]it moves with uniform velocity of 50 ft/sec. (\[{{t}_{2}}=10\,\sec \]acceleration is zero)
\[\therefore \] Distance covered,\[{{S}_{2}}=50\times 10=500ft.\]
(iii) Between 20s to 30s, acceleration is constant i.e., \[-5ft/{{s}^{2}}.\]
At 20s, velocity is \[50ft/s;\,\,t=30-20=10s.\]
\[\therefore \]Distance covered, \[{{S}_{3}}=ut+\frac{1}{2}a{{t}^{2}}\]
\[=50\times 10+\frac{1}{2}(-5){{(10)}^{2}}=500-250=250ft\]
\[\therefore \]Total distance travelled in 30s
\[=250+500+250=1000ft.\]
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