Answer to Question #92452 in Classical Mechanics for James Walker

Question

A car is travelling on a road. The maximum velocity the car can attain is 24 m/s & the maximum deceleration is 4 ms^-2. If car starts from rest and comes to rest after traveling 1032 m in the shortest time of 56 s, the maximum acceleration that the car can attain is?

Answer 1

As the maximum acceleration rate is 4 ms^-2 and the maximum velocity is 24 m/s, the time required to stop is:

$$t_{stop}=\frac{v_{max}}{a_{dec}}=\frac{24}{4}=6 s$$

The corresponding deceleration distance is:

$$S_{dec}=\frac{v_{max}^2}{2a_{dec}}=\frac{24^2}{2*4}=72 m$$

Now, we know that the car was accelerating for 56-6=50 seconds, and for the distance of 1032-72=960 meters.

This gives us:

$$L_{accel}=\frac{a_{accel} t_{accel }^2}{2}$$

$$a_{accel}={\frac{2L_{accel}}{t_{accel}}}=\frac{2*960}{50^2}=0.552 m/s^2$$

So, the maximum acceleration of the car is 0.552 m/s^2