A car is moving on road with velocity of 36 km/h. Driver presses the accelerator. Class 9

A car is moving on road with velocity of 36 km/h. Driver presses the accelerator. Class 9

Question 1.

A car is moving on road with velocity of 36 km/h. Driver presses the accelerator for a time interval of 10 s to increase the velocity to 54 km/h and then presses the brake to stop the car in a time interval of 5 s. Find the average acceleration. Class 9

i) when accelerator was pressed

ii) When brakes were pressed

Answer:

When accelerator was pressed, time interval = 10 s

Initial velocity u = 36 km/h = 36 × [latex]\frac{1000}{60 \times 60}[/latex] = 10 m/s

Final velocity v = 54 km/h = 54 × [latex]\frac{1000}{60 \times 60}[/latex] = 15 m/s

average acceleration a = t, [latex]\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}_2-\mathrm{t}_1}[/latex] = 15 - 10/10 = 0.5 m/s²

numerical value of velocity is increasing in given time interval, the average acceleration is in the direction of velocity.

ii) When brakes were pressed, time interval = 5 s

Initial velocity u = 54 km/h = 54 × [latex]\frac{1000}{60 \times 60}[/latex] = 15 m/s

Final velocity v = 0

average acceleration a = [latex]\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}_2-\mathrm{t}_1}[/latex] = 0 - 15/10 = - 3 m/s²

numerical value of velocity is decreasing in given time interval, the average acceleration is opposite to the direction of velocity

Question 2.

A ball is dropped from height Its velocity at t = 0 s is u = 0. At t = 1 s, its velocity is 9.8 m/s, at t = 2 s velocity increases to 19.6 m/s. At t = 3 s, velocity increases to 29.4 m/s. Find the average acceleration of ball in every successive interval of a second. Class 9

Answer:

average acceleration between 0 s and 1 s is a = [latex]\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}_2-\mathrm{t}_1}[/latex] = 9.8 - 0/1 -- 0 = 9.8 m/s²

average acceleration between 1 s and 2 s is a = [latex]\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}_2-\mathrm{t}_1}[/latex] = 19.6 - 9.8/2 - 1 = 9.8 m/s²