A particle accelerates from rest at a constan | vedclass

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(B) Let the time taken for acceleration be $t_1$ and the time taken for deceleration be $t_2$. Given,total time $T = t_1 + t_2 = 4\, s$. The velocity-

Vedclass · Accepted Answer

$16\, m$

Vedclass · Answer

(B) Let the time taken for acceleration be $t_1$ and the time taken for deceleration be $t_2$. Given,total time $T = t_1 + t_2 = 4\, s$. The velocity-time graph for this motion is a triangle with base $T = 4\, s$ and height $v_{max} = 8\, m s^{-1}$. The distance travelled is equal to the area under the velocity-time graph. Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Area $= \frac{1}{2} 	imes 4\, s 	imes 8\, m s^{-1} = 16\, m$. Thus,the total distance travelled is $16\, m$.

$A$ particle accelerates from rest at a constant rate for some time and attains a constant velocity of $8\, m s^{-1}$. Afterwards,it decelerates at a constant rate and comes to rest. If the total time taken is $4\, s$,the distance travelled is:

Similar Questions

How can you find the following?
$(i)$ Velocity from a displacement-time graph.
$(ii)$ Acceleration from a velocity-time graph.
$(iii)$ Displacement from a velocity-time graph.
$(iv)$ Velocity from an acceleration-time graph.

The following displacement-time graph shows the positions of a body at different times. Calculate the velocity of the body as it moves from: $(i) A-B, (ii) B-C, (iii) C-D$.

What is the quantity which is measured by the area occupied below the velocity-time graph?

What conclusion can you draw from the displacement$-$time graph of a body shown below?

The displacement-time graph of a body is parallel to the time axis. What will you infer about the velocity of the body?

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