Question

7. The acceleration of a cart started at t=0, varies with
time as shown in figure (3-E2). Find the distance
travelled in 30 seconds and draw the position-time graph.​

Answer 1

It is clear from the graph that the cart has uniform acceleration of 5 ft/s² for initial 10 s then uniform velocity for next 10 s (because acceleration is zero) and then retardation in further next 10 s. Area between graph and time axis gives the change in velocity in the taken time interval Let us consider three parts of each 10 s. In the first 10 s change of velocity = 5 ft/s² x 10 s = 50 ft/s, Since the cart starts from rest at t=0, average velocity in this period = (50+0)/2 =25 ft/s, Distance travelled in this 10 s =25 ft/s x 10 s =250 ft. From t=10 s to 20 s area under the graph is zero, so change in velocity is zero, it means that the velocity at t= 10 s (50 ft/s) remains unchanged. So distance travelled in this period = 50 ft/s x 10 s = 500 ft. From t=20 s to 30 s area under the graph = 5 ft/s² x 10 s = 50 ft/s but it is below the time axis, So this change of velocity is negative. It means that the cart moving with a velocity of 50 ft/s at t= 20 s comes to rest at t=30 s. So average velocity in this period = 25 ft/s and distance travelled = 25 ft/s x 10 s = 250 ft. So the distance travelled = 250 ft + 500 ft+ 250 ft = 1000 ft.Read more on Sarthaks.com - https://www.sarthaks.com/41444/the-acceleration-of-a-cart-started-at-t-0-varies-with-time-as-shown-in-figure-3-e2

Answer 2

The total distance traveled by the cart in 30 seconds is equal to 500 feet.

Given:

The acceleration for the first 10 seconds (a₁) = 5 ft s⁻²

The acceleration from t = 20 to t = 30 seconds (a₂) = -5 ft s⁻²

Total time period (T) = 30 seconds

To Find:

The distance traveled by the cart in 30 seconds.

Solution:

→ Since the acceleration of the cart is not constant, it is varying from time to time. Therefore we will divide the total journey of the cart into small three sub-intervals and calculate the individual distance traveled during these time intervals.

(i) t = 0 to t = 10 seconds:

Acceleration of the block (a₁) = 5 ft s⁻²

Time (t) = 10 seconds

Initial velocity of the block (u) = 0

→ We can calculate the distance traveled using the Second Equation of motion:

                                $\because S=ut+\frac{1}{2} at^{2} \\\\\therefore d_{1} =0+\frac{1}{2} \times (5)\times(10^{2} )\\\\\therefore d_{1} =\frac{500}{2}=250\:feet$

Therefore distance traveled (d₁) by the cart from t = 0 to t = 10 seconds is equal to 250 feet.

→ We can now calculate the final velocity of the block using the First Equation of motion:

                                 $\because v =u+at\\\\\therefore v=0+5\times10\\\\\therefore v=50\:ft\:s^{-2}$

→ The final velocity of the block at t = 10 second is equal to 50 ft s⁻¹.

(ii) t = 10 to t = 20 seconds:

Acceleration of the block (a) = 0 ft s⁻²

The initial velocity of the block (u) = 50 ft s⁻¹

Time (t) = 10 seconds

→ We can calculate the distance traveled using the Second Equation of motion:

                                $\because S=ut+\frac{1}{2} at^{2} \\\\\therefore d_{2} =50\times10+\frac{1}{2} \times (0)\times(10^{2} )\\\\\therefore d_{2} =500\:ft$

Therefore distance traveled (d₂) by the cart from t = 0 to t = 10 seconds is equal to 500 feet.

→ The final velocity (v) of the cart at t = 20 seconds will be equal to the initial velocity (u) of the cart which is 50 ft s⁻¹ because the acceleration is equal to zero.

(iii) t = 20 to t = 30 seconds:

Acceleration of the block (a₂) = -5 ft s⁻² (Retardation)

Time (t) = 10 seconds

The initial velocity of the block (u) = 50 ft s⁻¹

→ We can calculate the distance traveled using the Second Equation of motion:

                                $\because S=ut+\frac{1}{2} at^{2} \\\\\therefore d_{3} =(50)\times(10)+\frac{1}{2} \times (-5)\times(10^{2} )\\\\\therefore d_{3} =500-\frac{500}{2}\\\\\therefore d_{3} =250\:feet$

→ Therefore distance traveled (d₃) by the cart from t = 20 to t = 30 seconds is equal to 250 feet.

→ The total distance traveled by the cart will be equal to the sum of d₁,d₂, and d₃.

  ∴ Total distance traveled (d) = d₁ + d₂ + d₃.

                                            ∴ d = (250) + (500) + (250)

                                            ∴ d = 1000 feet

Hence the total distance traveled by the cart in 30 seconds is equal to 500 feet.

→ The following will be the Position-Time graph of the journey:

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