Question

The velocity time graph of a car is given below. The car weighs 1000 kg. (i) What is the distance travelled by the car in first 2 seconds?
(ii) What is the braking force at the end of 5 seconds to bring the car to a stop within one second?
b) Derive the equation S=ut+1/2 at² using graphical method.

Answer 1

(i) The distance travelled by the car is first 2 seconds is 15 m

(ii) The braking force at the end of 5 seconds to bring the car to a stop within one second is 15000 N

Explanation:

The graph is attached

Given mass of car m = 1000 kg

(a) We know that in a velocity time graph, the distance travelled in time t = area under the velocity-time graph upto time t

Therefore, the distance travelled by the car in first 2 seconds

= Area of ΔABE

$=\frac{1}{2}\times BE\times AE$

$=\frac{1}{2}\times 15\times 2$

$=15$ m

(b) From graph it is clear that the car moves with a constant acceleration from A to B and then with a zero acceleration from B to C and then with negative acceleration from C to D

Acceleration at the end of 5 seconds

= Acceleration (a) from  C to D = Slope of line CD

                                                = $\frac{15-0}{5-6}$

                                                = $=-15$ m/s²

Therefore the breaking force at the end of 5 seconds

$F=ma$

$\implies F=1000\times (-15)$

$\implies F=-15000$ N

(negative sign indicates that force applied is in the opposite direction of motion)

(b) Distance travelled from C to D

s = Area of ΔCFD

$=\frac{1}{2}\times CF\times FD$

$=\frac{1}{2}\times 15\times (6-5)$

$=\frac{15}{2}$ m

We have calculated, acceleration a from velocity C to velocity D as -15 m/s²

Initial velocity at C, u = 15 m/s

Time interval t = 6-5 = 1 seconds

Therefore,

$ut+\frac{1}{2}at^2$

$=15\times 1+\frac{1}{2}(-15)\times 1^2$

$=15-\frac{15}{2}$

$=\frac{15}{2}$

$=s$

Hope this answer is helpful.

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