Physics, asked by subhoshsingh9835, 11 months ago


59. A car is moving on a straight road with uniform
acceleration. The speed of the car varies with time as
follows:
Time (s) : 0 2 4 6 8 10
Speed (m/s) : 4 8 12 16 20 24
Draw the speed-time graph by choosing a convenient
scale. From this graph:
Calculate the acceleration of the car.
i) Calculate the distance travelled by the car in 10 second​

Answers

Answered by Acekiller
14

so the scale will be

on x axis 1cm=2sec

on y axis 1cm = 4 m/s

acceleration of car is 2m/s²

distance covered in 10ssec is 140 m

Attachments:
Answered by mudit2600
0

(i) Acc(i) Acceleration of the car = slope of AB=

10−0

24−4

=

10

20

=2m/s

2

(ii) Distance travelled by the car in 10s= area of trapezium OABC

=

2

1

×(OA+BC)×OC

=

2

1

×(4+24)×10=140m

solutioneleration of the car = slope of AB=

10−0

24−4

=

10

20

=2m/s

2

(ii) Distance travelled by the car in 10s= area of trapezium OABC

(i) Acceleration of the car = slope of AB=

10−0

24−4

=

10

20

=2m/s

2

(ii) Distance travelled by the car in 10s= area of trapezium OABC

=

2

1

×(OA+BC)×OC

=

2(i) Acceleration of the car = slope of AB=

10−0

24−4

=

10

20

=2m/s

2

(ii) Distance travelled by the car in 10s= area of trapezium OABC

=

2

1

×(OA+BC)×OC

=

2

1(i) Acceleration of the car = slope of AB=

10−0

24−4

=

10

20

=2m/s

2

(ii) Distance travelled by the car in 10s= area of trapezium OABC

=

2

1

×(OA+BC)×OC

=

2

1

×(4+24)×10=140m

solution

×(4+24)×10=140m

solution

1

×(4+24)×10=140m

solution

=

2

1

×(OA+BC)×OC

=

2

1

×(4+24)×10=140m

solution

Answer:

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A car is moving on a straight road with uniform acceleration. The speed of the car varies with time as follows :

Time(s): 0 2 4 6 8 10

Speed (m/s): 4 8 12 16 20 24

Draw the speed-time graph by choosing a convenient scale. From this graph :

(i) Calculate the acceleration of car.

(ii) Calculate the distance travelled by the car in 10 seconds.

Medium

Solution

verified

Verified by Toppr

(i) Acceleration of the car = slope of AB=

10−0

24−4

=

10

20

=2m/s

2

(ii) Distance travelled by the car in 10s= area of trapezium OABC

=

2

1

×(OA+BC)×OC

=

2

1

×(4+24)×10=140m

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