Answer to Question #187287 in Calculus for John

Question

A car has an initial velocity of 20 m s⁻¹

For the first 4 seconds of its motion it accelerates at 2.5 m·s^{−2 For} the next T seconds it travels at a constant velocity of V m s⁻¹. The car then decelerates to rest

Sketch the velocity-time graph for the whole journey of the car. 

Find V.

The total time for the journey is 40 seconds.

 

iii) If the total distance travelled by the car is 1090m, find T.

Answer 1

Case(i) Initial velocity u= 20 m/s

time period $t_1=4 \ s$

Acceleration = 2.5 m/s$^2$

So, Final velocity $v_1=u_1+a_1t_1$

$= 20+(2.5\times4)\\=30\ m/s$

case (ii) Initial velocity of car = 30 m/s

Time period = T sec

Final velocity =30 m/s

case(iii)

Time period =T'

So, $v_2=u_2+a_2t_2$

$0=30+a_2\cdot T'\\a_2=-\dfrac{30}{T'}$

(i)So V= $v_1=u_2=30\ m/s$

Total time duration = 40 s

and total distance travelled = 1090 m

So, distance covered in remaining 36 sec= 990m

Distance covered when car is in constant speed = $30T$

Distance covered during deacceleration :

$s_2 = u_2(36-T)+\frac{1}{2}a_2(36-T)^2\\990=30(36-T)+\frac{1}{2}(-\frac{5}{11})(36-T)^2\\$

On solving

we get T= 30 sec