The velocity-time graph for a vehicle is as shown below figure. Draw the acceleration-time graph from it.
A driver of a car travelling at \(52kmh^{-1}\) applies the breaks and deaccelerates uniformly. The car stops in \(5s\), another driver going at \(34kmh^{-1}\) applies his breaks slower and stops after \(10sec\). Plot the speed vs time graph for two cars. Which of the two cars travelled further after the breaks were applied?
Ans: \(2^{nd}\) car travelled farther than the first car after the brakes were applied.
A train moves from one station to another in two hours time. Its speed vs time graph is shown in below figure. (i) Determine the maximum acceleration during the journey? (ii) Calculate: the distance covered during the time interval from \(0.75 hours\) to \(1 hours\).
Ans:(i) \(a_{max}=120km.h^{-2}\). (ii) \(8.75km\).
A motor car starts from rest, moves with a uniform acceleration and attains velocity of \(8ms^{-1}\) in \(8s\). It is then move with a uniform velocity and finally brought to rest in \(32m\) under uniform retardation. The total distance covered by car is \(464m\).Find: (i) Acceleration. (ii) Retardation. and (iii) Total Time Taken.
Ans:(i) \(a=1m.s^{-2}\). (ii) \(a=-1m.s^{-2}\). and (iii) \(t_{total}=66Sec\).
A ball is thrown upwards with an initial velocity of \(10ms^{-1}\). After how much time will it return? Draw the velocity-time graph for the ball and find from the graph: (i) The maximum height attained by the ball? and (ii) Height of the ball after \(15s\). TAKE: \(g=10ms^{-2}\).
Ans:(i) \(H_{max}=500m\). and (ii) \(H_{15s}=375m\).
The position-time graph of a particle moving along the straight line is shown below. Draw: (i) Velocity-Time Graph. (ii) Acceleration-Time Graph.
A particle starts from rest at time \(t=0\) and moves along a straight line with an acceleration \(a\) \(ms^(-2)\) a shown below. Find: (a) the time at which the speed of the particle is maximum. (b) Also; calculate the displacement of the particle from starting point.
Ans:(a) t=2s and (b) \(S=-40m\).
A car accelerates from rest at a constant rate A for sometimes and after which it retards at constant rate B to come to rest. If the total time lapse is T seconds, evaluate: the maximum velocity reached and total distance travelled in terms of A.
Ans: \(v_{max}=\left(\frac{AB}{A+B}\right)T\). and \(\left(\frac{AB}{A+B}\right)\frac{T^{2}}{2}\).
The acceleration-time graph for a particle in rectilinear motion is shown below. Find: the average acceleration in first 10 Seconds? Ans: \(15m.s^{-2}\).
