Question

The velocity-time graph of a car is given below. The car weigh 1000 is the car in the 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

Answer 1

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

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

Explanation:

Note: The question in the images is entirely different. The answer here provided is based on the question asked.

The graph for this question is attached.

Given mass of SUV 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 SUV in first 2 seconds

= Area of ?ABE

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

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

$=15$ m

(b) From graph it is clear that the SUV 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)

Hope this answer is helpful.

Know More:

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