The v-t graph for a particle is as shown. The distance travelled in the first four second is
Area bounded by v-t graph represents displacement. Here.
Area = (1/2)×4×8= 16
So, Distance travelled is 16 m.
A particle starts form rest. Its acceleration (a) versus time (t) is as shown in the figure. The maximum speed of the particle will be
Area of acceleration time graph represents change in velocity. As acceleration is always positive speed is always increasing.
Area = (1/2)× 10× 11 = 55
So, max. speed of 55 m/s
A rocket is fired upwards. Its velocity versus time graph is shown in figure. The maximum height reached by the rocket is :
The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are as shown in the figure. Choose the incorrect statement regarding these graphs :
Slope of x-t graph represents velocity. slope is more for B so it will have higher velocity
A car accelerates from rest at a constant rate \(\alpha\) for some time after which it decelerates at a constant rate \(\beta\) and comes to rest. If total time elapsed is t, then maximum velocity acquired by car will be:
(1994)
Let (v_{max}) be the maximum velocity. Time to accelerate: \(t_1 = \frac{v_{max}}{\alpha}). Time to decelerate: (t_2 = \frac{v_{max}}{\beta}). Total time (t = t_1 + t_2 = \frac{v_{max}}{alpha} + \frac{v_{max}}{beta} = v_{max}\left(\frac{1}{alpha} + \frac{1}{\beta}\right) = v_{max}\left(\frac{\beta + \alpha}{\alpha\beta}\right)). Solving for \(v_{max}): (v_{max} = \frac{\alpha\beta t}{\alpha + \beta}).