Figure shows the graph of acceleration of particle as a function of time. The maximum speed of the particle is (particle starts from rest)

Question

Figure shows the graph of acceleration of particle as a function of time. The maximum speed of the particle is (particle starts from rest) (A) 7 m/s (B) 8 m/s (C) 4 m/s (D) 16 m/s

Tardigrade Admin · Accepted Answer

Maximum speed will occur at t = 4 s, because till then a is positive. Now change in velocity is area under acceleration time graph, so v4s - 0 = (1/2) [ 4× 4] = 8 ⇒ v4s = 8 m/s

Q. Figure shows the graph of acceleration of particle as a function of time. The maximum speed of the particle is (particle starts from rest)

7 m/s

8 m/s

4 m/s

16 m/s

Maximum speed will occur at $t = 4 s,$ because till then a is positive. Now change in velocity is area under acceleration time graph, so

$v_{4 s} - 0 = \frac{1}{2} [4 \times 4] = 8$

$\Rightarrow v_{4 s} = 8 m / s$

Q. Figure shows the graph of acceleration of particle as a function of time. The maximum speed of the particle is (particle starts from rest)

7 m/s

8 m/s

4 m/s

16 m/s

Maximum speed will occur at t=4s, because till then a is positive. Now change in velocity is area under acceleration time graph, so v4s​−0=21​[4×4]=8 ⇒v4s​=8m/s

Maximum speed will occur at $t = 4 s,$ because till then a is positive. Now change in velocity is area under acceleration time graph, so

$v_{4 s} - 0 = \frac{1}{2} [4 \times 4] = 8$

$\Rightarrow v_{4 s} = 8 m / s$